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New Zealand
Level 8 - NCEA Level 3
# Applications of Pythagorean identities
Lesson
The fundamental identity $\sin^2\theta+\cos^2\theta\equiv1$sin2θ+cos2θ1 is used in simplifying expressions and in establishing further trigonometric identities.
For example, we may wish to verify that $1+\tan^2\theta=\sec^2\theta$1+tan2θ=sec2θ, a formula that occurs in certain problems in calculus. A strategy for this is to manipulate one side of the equation by means of the fundamental identity until it can be seen to be the same as the other side. In this case we might write
$LHS$LHS $=$= $1+\tan^2\theta$1+tan2θ $=$= $1+\frac{\sin^2\theta}{\cos^2\theta}$1+sin2θcos2θ $=$= $\frac{\cos^2\theta}{\cos^2\theta}+\frac{\sin^2\theta}{\cos^2\theta}$cos2θcos2θ+sin2θcos2θ $=$= $\frac{\cos^2\theta+\sin^2\theta}{\cos^2\theta}$cos2θ+sin2θcos2θ $=$= $\frac{1}{\cos^2\theta}$1cos2θ $=$= $\sec^2\theta$sec2θ $=$= $RHS$RHS
#### Example 1
Simplify the expression $\left(\cos\beta+1\right)\left(\cos\beta-1\right)$(cosβ+1)(cosβ1) and write it in terms of the sine function.
Expanding the brackets gives $\cos^2\beta-1$cos2β1. But this is $-\left(1-\cos^2\beta\right)$(1cos2β) and hence it is equivalent to $-\sin^2\beta$sin2β.
#### Example 2
Given that $x=4\sin\theta$x=4sinθ and $y=5\cos\theta$y=5cosθ, find a relation between $x$x and $y$y that does not involve the trigonometric functions.
In order to use the Pythagorean identity we need the squares of the trigonometric functions. So, we write $x^2=16\sin^2\theta$x2=16sin2θ and $y^2=25\cos^2\theta$y2=25cos2θ. Then, $\frac{x^2}{16}=\sin^2\theta$x216=sin2θ and $\frac{y^2}{25}=\cos^2\theta$y225=cos2θ. On adding these two equations we obtain $\frac{x^2}{16}+\frac{y^2}{25}=\sin^2\theta+\cos^2\theta$x216+y225=sin2θ+cos2θ, and finally
$\frac{x^2}{16}+\frac{y^2}{25}=1$x216+y225=1
This equation fixes the points $\left(x,y\right)$(x,y) that describe an ellipse in the coordinate plane. Evidently, the same ellipse can be specified by the two trigonometric functions in the parameter $\theta$θ.
#### More Worked Examples
##### QUESTION 1
Simplify the expression $\frac{1}{1-\sin x}\times\frac{1}{1+\sin x}$11sinx×11+sinx.
##### QUESTION 2
Prove the identity $2\cos^2\left(\theta\right)-3=-1-2\sin^2\left(\theta\right)$2cos2(θ)3=12sin2(θ).
##### QUESTION 3
If $x=4\sin\theta$x=4sinθ and $y=3\cos\theta$y=3cosθ, form an equation relating $x$x and $y$y that does not involve $\sin\theta$sinθ or $\cos\theta$cosθ.
### Outcomes
#### M8-6
Manipulate trigonometric expressions
#### 91575
Apply trigonometric methods in solving problems |
# Difference between revisions of "2018 AMC 12B Problems/Problem 7"
## Problem
What is the value of $$\log_37\cdot\log_59\cdot\log_711\cdot\log_913\cdots\log_{21}25\cdot\log_{23}27?$$ $\textbf{(A) } 3 \qquad \textbf{(B) } 3\log_{7}23 \qquad \textbf{(C) } 6 \qquad \textbf{(D) } 9 \qquad \textbf{(E) } 10$
## Solution 1
From the Change of Base Formula, we have $$\frac{\prod_{i=3}^{13} \log (2i+1)}{\prod_{i=1}^{11}\log (2i+1)} = \frac{\log 25 \cdot \log 27}{\log 3 \cdot \log 5} = \frac{(2\log 5)\cdot(3\log 3)}{\log 3 \cdot \log 5} = \boxed{\textbf{(C) } 6}.$$
## Solution 2
Using the chain rule of logarithms $\log _{a} b \cdot \log _{b} c = \log _{a} c,$ we get \begin{align*} \log_37\cdot\log_59\cdot\log_711\cdot\log_913\cdots\log_{21}25\cdot\log_{23}27 &= (\log _{3} 7 \cdot \log _{7} 11 \cdots \log _{23} 27) \cdot (\log _{5} 9 \cdot \log _{9} 13 \cdots \log _{21} 25) \\ &= \log _{3} 27 \cdot \log _{5} 25 \\ &= 3 \cdot 2 \\ &= \boxed{\textbf{(C) } 6}. \end{align*}
~ pi_is_3.14
## See Also
2018 AMC 12B (Problems • Answer Key • Resources) Preceded byProblem 6 Followed byProblem 8 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 All AMC 12 Problems and Solutions
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.
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# Find the distance and midpoint for the following two points. (5,8), (-3,2)
txmedteach | Certified Educator
To start, we need to think of what distance and midpoint mean in this context. Let's graph these two points to see what we're looking at.
To find the midpoint is not that bad, actually! We simply take the average x value and average y value! Here's the formula:
`M = ((x_1+x_2)/2, (y_1+y_2)/2)`
All we need to do here is substitute our x and y values into the formula:
`M = ((5+(-3))/2, (8+2)/2)`
Now, we simplify to get:
`M = (1, 5)`
Let's graph the point to see if we're right.
Well, that certainly looks like the midpoint to me! If you want to check, you can see that the midpoint is 3 squares above and 4 squares to the right of (-3, 2), and to get to (5,8) from our midpoint, we need to move the same 3 up and 4 to the right. It is confirmed!
Now, to find the distance. The distance formula proposed is useful (shown here):
`D = sqrt((y_2-y_1)^2 + (x_2-x_1)^2)`
However, this doesn't really hit at the basis for how we find distance. If you wanted to find distance, you can treat your points like points on a right triangle. To complete the triangle, make a vertical line from one point, and a horizontal line from another:
Recognize that the hypotenuse of this right triangle is also the distance between the two points! Also, recall the Pythagorean Theorem:
`a^2 + b^2 = c^2`
So, considering that our hypotenuse is c, all we need to do is find the length of the legs to solve this equation! If you look at the graph, you'll easily see that the leg lengths are 8 for the horizontal leg and 6 for the vertical leg.
In the distance formula, this is represented by the differences between x and y terms, so you'll see this way is no different!
`x_2-x_1 = -3-5 = -8`
`y_2 - y_1 = 2-8 = -6`
Well, -6 is different from 6, same with -8 and 8. However, since we're taking the square of the values, there isn't a difference!
We'll put these values into the Pythagorean Theorem/Distance Formula:
`64 + 36 = c^2`
`100 = c^2`
Now, we take the square root:
`10 = c`
So our distance will be 10. Another way to see this is that the triangle we created on the graph is a 3,4,5 right triangle (just doubled).
I hope that helps! |
# Math Expressions Grade 3 Student Activity Book Unit 1 Lesson 12 Answer Key Multiply and Divide with 4
This handy Math Expressions Grade 3 Student Activity Book Answer Key Unit 1 Lesson 12 Multiply and Divide with 4 provides detailed solutions for the textbook questions.
## Math Expressions Grade 3 Student Activity Book Unit 1 Lesson 12 Multiply and Divide with 4 Answer Key
Explore Patterns with 4s
What patterns do you see below?
Inter relation between addition and multiplication by increase in integral value,
Explanation:
In the above pattern we can see 4 table in the form of multiplication and the expanded form of it in the form of addition, example in 8 = 2 × 4 the number 4 if added 2 times we get 8, in similar way as 12 = 3 × 4 if 4 is added 3 times we get 12, hence we can say the addition of a number is based on the integral value such as 1, 2, 3 and so on.
Use the 5s Shortcut for 4s
Solve each problem.
Question 1.
How many legs are on 6 horses? Find the total by starting with the fifth count-by and counting up from there.
There are total of 24 legs in 6 horses,
Explanation:
First we need to make a group of 5 horses and then we can find that 1 horse is remaining, the number of legs in first 5 horses = 5 × 4 = 20, Number of legs in the remaining horses = 1 × 4 = 4, Total number of sides = 20 + 4 = 24.
Question 2.
How many sides are in 8 quadrilaterals? Find the total by starting with the fifth count-by and counting up from there.
This large rectangle is made up of two small rectangles.
There are total of 32 sides in 8 quadrilaterals,
Explanation:
The given quadrilaterals have 4 sides, First we need to make a group of 5 quadrilaterals and then we can find that 3 quadrilaterals are remaining, the number of sides in first 5 quadrilaterals = 5 × 4 = 20, Number of sides in the remaining quadrilaterals = 3 × 4 = 12, Total number of sides = 20 + 12 = 32.
Question 3.
Find the area of the large rectangle by finding the areas of the two small rectangles and adding them.
The area of the large rectangle is 28 square units,
Explanation:
Green rectangle: length = 5 units, breadth = 4 units, Yellow rectangle: length = 4 units, breadth = 2 units,
In the large rectangle given below there are two small rectangles, Area of the large rectangle = Area of green rectangle + Area of yellow rectangle = length × breadth + length × breadth = 5 × 4 + 4 × 2 = 20 + 8 = 28 square units.
Question 4.
Find the area of the large rectangle by multiplying the number of rows by the number of square units in each row.
The area of the large rectangle is 28 square units,
Explanation:
Number of rows in the large rectangle = 7, Number of squares in each row = 4, Area of the large rectangle = Number of rows in the large rectangle × Number of squares in each row = 7 × 4 = 28 square units.
Use Multiplications You Know
You can combine multiplications to find other multiplications.
This Equal Shares Drawing shows that 7 groups of 4 is the same as 5 groups of 4 plus 2 groups of 4.
Question 5.
5 × 4 = 20, 2 × 4 = 8, adding the answers we get 7 × 4 = 20 + 8 = 28,
Explanation:
5 × 4 = 5 + 5 + 5 + 5 = 20,
As the number to be added is 5 and the integral number value is 4, so to find the value we need to add 5 for 4 times, 5 + 5 + 5 + 5 = 20,
2 × 4 = 2 + 2 + 2 + 2 = 8,
As the number to be added is 2 and the integral number value is 4, so to find the value we need to add 2 for 4 times, 2 + 2 + 2 + 2 = 8,
To add the answers we need to combine multiplications 5 × 4 and 2 × 4, we need to add the integral values 5 and 2 then we get 7 × 4 = 5 × 4 + 2 × 4 = 20 + 8 = 28.
Question 6.
Find 7 × . Did you get the same answer as in exercise 5?
7 × 4 = 28, Yes we get the same answer as in exercise 5,
Explanation:
7 × 4 = 7 + 7 + 7 + 7 = 28,
As the number to be added is 7 and the integral number value is 4, so to find the value we need to add 7 for 4 times, 7 + 7 + 7 + 7 = 28.
Question 7.
Find this product: 5 × 4 = ____
5 × 4 = 5 + 5 + 5 + 5 = 20,
Explanation:
As the number to be added is 5 and the integral number value is 4, so to find the value we need to add 5 for 4 times, 5 + 5 + 5 + 5 = 20.
Question 8.
Find this product: 4 × 4 = ____
4 × 4 = 4 + 4 + 4 + 4 = 16,
Explanation:
As the number to be added is 4 and the integral number value is 4, so to find the value we need to add 4 for 4 times, 4 + 4 + 4 + 4 = 16.
Question 9.
Use your answers to exercises 7 and 8 to find this product: 9× 4 = ____
9 × 4 = 9 + 9 + 9 + 9 = 36,
Explanation:
To find 9 × 4 we need to combine multiplications 5 × 4 and 4 × 4, we need to add the integral values 5 and 4 then we get 9 × 4 = 5 × 4 + 4 × 4 = 20 + 16 = 36.
Question 10.
Make a drawing to show that your answers to exercises 7-9 are correct.
Explanation:
9 × 4 = 9 + 9 + 9 + 9 = 36,
To find 9 × 4 we need to combine multiplications 5 × 4 and 4 × 4, we need to add the integral values 5 and 4 then we get 9 × 4 = 5 × 4 + 4 × 4 = 20 + 16 = 36.
What’s the Error?
Dear Math Students,
Today I had to find 8 × 4. I didn’t know the answer, but I figured it out by combining two multiplications I did know:
Is my answer right? If not, please correct my work and tell me why it is wrong.
The Puzzled Penguin
Question 11.
Write an answer to the Puzzled Penguin.
The answer given by Penguin for 8 × 4 is wrong,
Explanation:
We can solve by combining multiplications of 5 × 2 and 3 × 2, for this we need to add only the integral values 5 and 3 but not 2s, then we get 8 × 2 = 5 × 2 + 3 × 2 = 10 + 6 = 16.
Make Sense of Problems
Write an equation and solve the problem.
Question 12.
Galen has 20 pictures to place in his book. If he puts 4 pictures on each page, how many pages will he fill?
5 pages were filled by Galen with 4 pictures in each,
Explanation:
Total number of pictures Galen has = 20, Number of pictures placed in each page = 4, Number of pages filled = $$\frac{Total number of pictures Galen has }{Number of pictures placed in each page}$$ = $$\frac{20}{4}$$ = 5.
Question 13.
Emery arranged the tiles in an array with 4 columns and 7 rows. How many tiles were in the array?
Emery arranged 28 tiles in an array,
Explanation:
Number of columns = 4, Number of rows = 7, Total number of tiles in the array = Number of rows × Number of columns = 7 × 4 = 28. |
# Rational Exponents, Radicals, and Complex Numbers
## Presentation on theme: "Rational Exponents, Radicals, and Complex Numbers"— Presentation transcript:
Rational Exponents, Radicals, and Complex Numbers
Rational Exponents, Radicals, and Complex Numbers
CHAPTER 8 Rational Exponents, Radicals, and Complex Numbers 8.1 Radical Expressions and Functions 8.2 Rational Exponents 8.3 Multiplying, Dividing, and Simplifying Radicals 8.4 Adding, Subtracting, and Multiplying Radical Expressions 8.5 Rationalizing Numerators and Denominators of Radical Expressions 8.6 Radical Equations and Problem Solving 8.7 Complex Numbers
8.2 Rational Exponents 1. Evaluate rational exponents.
2. Write radicals as expressions raised to rational exponents. 3. Simplify expressions with rational number exponents using the rules of exponents. 4. Use rational exponents to simplify radical expressions.
Rational exponent: An exponent that is a rational number.
Rational Exponents with a Numerator of 1 a1/n = where n is a natural number other than 1. Note: If a is negative and n is odd, then the root is negative. If a is negative and n is even, then there is no real number root.
Example Rewrite using radicals, then simplify if possible. a. 491/2 b. 6251/4 c. (216)1/3 Solution a. b. c.
continued Rewrite using radicals, then simplify. d. (16)1/4 e. 491/2 f. y1/6 Solution d. e. f.
continued Rewrite using radicals, then simplify. g. (100x8)1/2 h. 9y1/5 i. Solution d. e. f.
General Rule for Rational Exponents
where a 0 and m and n are natural numbers other than 1.
Example Rewrite using radicals, then simplify, if possible.
a. 272/3 b. 2433/5 c. 95/2 Solution a. b. c.
continued Rewrite using radicals, then simplify, if possible. d. e. f.
Solution d. e. f.
Negative Rational Exponents
where a 0, and m and n are natural numbers with n 1.
Example Rewrite using radicals; then simplify if possible.
a. 251/2 b. 272/3 Solution a. b.
continued Rewrite using radicals; then simplify if possible. c. d.
Solution c.
Example Write each of the following in exponential form. a. b.
Solution a. b.
continued Write each of the following in exponential form. c. d.
Solution c. d.
Rules of Exponents Summary
(Assume that no denominators are 0, that a and b are real numbers, and that m and n are integers.) Zero as an exponent: a0 = 1, where a 0. 00 is indeterminate. Negative exponents: Product rule for exponents: Quotient rule for exponents: Raising a power to a power: Raising a product to a power: Raising a quotient to a power:
Example Use the rules of exponents to simplify. Write the answer with positive exponents. Solution Use the product rule for exponents. (Add the exponents.) Add the exponents. Simplify the rational exponent.
Example Use the rules of exponents to simplify. Write the answer with positive exponents. Solution Use the product rule for exponents. (Add the exponents.) Rewrite the exponents with a common denominator of 6. Add the exponents.
Example Use the rules of exponents to simplify. Write the answer with positive exponents. Solution Use the quotient for exponents. (Subtract the exponents.) Rewrite the subtraction as addition. Add the exponents.
Example Use the rules of exponents to simplify. Write the answer with positive exponents. Solution Add the exponents.
Example Use the rules of exponents to simplify. Write the answer with positive exponents. Solution
Example Use the rules of exponents to simplify. Write the answer with positive exponents. Solution
Example Use the rules of exponents to simplify. Write the answer with positive exponents. Solution
Example Rewrite as a radical with a smaller root index. Assume that all variables represent nonnegative values. a. b. Solution
continued Rewrite as a radical with a smaller root index. Assume that all variables represent nonnegative values. c. Solution
Example Perform the indicated operations. Write the result using a radical. a. b. Solution a. b.
continued Perform the indicated operations. Write the result using a radical. c. Solution c.
Example Write the expression below as a single radical. Assume that all variables represent nonnegative values. Solution |
# How do you find int (x+1)/(x(x^2-1)) dx using partial fractions?
Dec 30, 2015
You try to split the rational function into a sum that will be really easy to integrate.
#### Explanation:
First of all : ${x}^{2} - 1 = \left(x - 1\right) \left(x + 1\right)$.
Partial fraction decomposition allows you to do that :
$\frac{x + 1}{x \left({x}^{2} - 1\right)} = \frac{x + 1}{x \left(x - 1\right) \left(x + 1\right)} = \frac{1}{x \left(x - 1\right)} = \frac{a}{x} + \frac{b}{x - 1}$ with $a , b \in \mathbb{R}$ that you have to find.
In order to find them, you have to multiply both sides by one of the polynomials at the left of the equality. I show one example to you, the other coefficient is to be found the same way.
We're gonna find $a$ : we have to multiply everything by $x$ in order to make the other coefficient disappear.
$\frac{1}{x \left(x - 1\right)} = \frac{a}{x} + \frac{b}{x - 1} \iff \frac{1}{x - 1} = a + \frac{b x}{x - 1}$.
$x = 0 \iff - 1 = a$
You do the same thing in order to find $b$ (you multiply everything by $\left(x - 1\right)$ then you choose $x = 1$), and you find out that $b = 1$.
So $\frac{x + 1}{x \left({x}^{2} - 1\right)} = \frac{1}{x - 1} - \frac{1}{x}$, which implies that $\int \frac{x + 1}{x \left({x}^{2} - 1\right)} \mathrm{dx} = \int \left(\frac{1}{x - 1} - \frac{1}{x}\right) \mathrm{dx} = \int \frac{\mathrm{dx}}{x - 1} - \int \frac{\mathrm{dx}}{x} = \ln \left\mid x - 1 \right\mid - \ln \left\mid x \right\mid$ |
Simple equations on a graph
## Start by thinking of a number line
When we learn to add and subtractmultiply and divide, we think of numbers as being all along the number line. But in algebra we think about numbers as being in a plane.
## Another number line crosses it
Picture two number lines crossing each other at zero. The one that goes up and down we call the y-axis and the one that goes from side to side we call the x-axis. The x-axis is always perpendicular to the y-axis – at right angles to it.
## Naming points on a plane
All of the points on the plane can be described by naming a y number and an x number. So y=3, x=4 means you go up three and right four and make a point there. The negative numbers are down and to the left, so y = -3, x = -4 means you go down three and left four and make a point there. Where the two lines meet, that’s zero.
## An equation names a line
The equation y = 3 means that you draw all of the points on the number plane where y is equal to 3. So you make a point where x = 0 and y = 3, and another point where x = 1 and y = 3, and another point where x = 2 and y = 3, and so on. That gives you a horizontal line, three numbers above zero.
### What is a line?
The equation x = 3 means that you draw all of the points on the number plane where x is equal to 3. So you make a point where y = 0 and x = 3, and so on. That gives you a vertical line, three numbers to the right of zero. |
The Quotient Rule is an important formula for finding finding the derivative of any function that looks like fraction. It is just one of many essential derivative rules that you’ll have to master in order to succeed on the AP Calculus exams. In this article I’ll show you the Quotient Rule, and then we’ll see it in action in a few examples.
## The Quotient Rule
There are many different but equivalent ways to express the Quotient Rule.
Suppose f and g are differentiable functions. Then the quotient function, f/g is also differentiable, and
Or, in a more compact way (using letters u and v for the functions):
Finally, many textbooks give the following version using Leibniz notation. You may be familiar with the “Hi – Lo” trick that helps us to memorize it. “Hi” means the top (numerator) function, “Lo” is the bottom (denominator) function, and “De” tells you to take a derivative.
### Relationship to Product Rule
The Quotient Rule and Product Rule share much in common. In fact, the two formulas are similar if you look at them the right way. (Unfortunately, most textbooks that I’ve seen do not write these formulas in the way that highlights their similarities.)
ProductQuotient
f = uvf = u/v
f ' = u'v + uv'f ' = (u'v - uv')/v2
Notice that the quotient formula is just like the product formula except that there is a minus (-) in the middle instead of (+), and there is an added feature: denominator squared.
### Example
Find the derivative of .
Identify the top and bottom functions,
u = x2 + 3x – 1.
v = x4 + 2.
It may help to write down the derivatives of u and v separately:
u' = 2x + 3.
v' = 4x3.
Then, using the Quotient Rule formula as a recipe, we have all the ingredients to cook up the derivative!
Now this is a correct but unsimplified derivative. Often on the AP Calculus test, your answers may not match the answer choices, simply because they have given their answers in simplest form. Let’s simplify our answer by multiplying through on the numerator and collecting like terms.
In order to help you see the steps, I’ve colored terms from the first product in blue, and those from the second in red.
### Example: Avoiding Quotient Rule
As much as we love the quotient rule, sometimes it’s just overkill. Sometimes a function has the form of a fraction but can easily be simplified to a non-fractional form. Then other derivative rules might be used instead of quotient rule.
Find the slope of the curve at x = 1.
At first, this may look like a job for the Quotient Rule. However notice that the denominator has only a single term. The function can be simplified before taking the derivative.
Then take the derivative (because the derivative is the tool that measures slope).
Plug in x = 1 to find the slope: 8(1) – (1)-2 = 7.
## Conclusion
If a function has the form of a fraction, then you may use the Quotient Rule to find its derivative, but also be on the lookout for easy simplifications that avoid having to use the rule. |
What is the vertex of y=x^2 - 8x - 3 ?
Jan 1, 2016
The Solution set(or vertex set) is: $S = \left\{4 , - 19\right\}$
Explanation:
The general formula for a quadratic function is:
$y = A {x}^{2} + B x + C$
To find the vertex, we apply those formulas:
${x}_{v e r t e x} = - \frac{b}{2 a}$
${y}_{v e r t e x} = - \frac{\triangle}{4 a}$
In this case:
${x}_{v e r t e x} = - \frac{- 8}{2 \cdot 1} = - \left(- 4\right) = 4$ and
${y}_{v e r t e x} = - \frac{{b}^{2} - 4 a c}{4 \cdot 1} = - \frac{64 - 4 \cdot 1 \cdot \left(- 3\right)}{4}$
${y}_{v e r t e x} = - \frac{76}{4} = - 19$
So, the Solution set(or vertex set) is: $S = \left\{4 , - 19\right\}$ |
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# Introduction to Graphs | Class 8 Maths
A graph is a mathematical representation of networks. The purpose of the graph is to show mathematical relations in visual form so that it can be easily understood. There are many types of graph:
• Bar graph
• Pie graph
• Line graph
### Bar Graph
Bar graph, also known as bar chart which is a visual tool and is used to compare the data among categories. Bar graph may be horizontal or vertical.
e.g. Below is a bar graph which shows pollution level of each city.
Bar graph represents a discrete value on one axis and category on another axis and the motive is to show the relation between two axes.
The bar graph is an easy way to make a comparison between the set of data.
### Pie Chart
A pie chart is a circular chart in which each data is represented in a portion of circle. As the chart is divided into wedge-like sectors the total value of pie chart is always 100%.
e.g. The pie chart below is used to represent people’s choice of Laptop’s brands. The circle as a whole here is represented by all the people who took part in the survey. Since it is a whole, the sum of all percentages represented in a pie graph must add up to 100%.
#### Steps for Creating a Pie Chart
Step 1: Decide the topic of your chart.
Step 2: Having all the information or data and divide it into a number of items, and the value of each item adding together should have a sum equal to 100%.
When comparing parts of a whole, pie-chart is the ideal method.
### Histograms
Both histogram and bar graph are similar but there is a difference between the two that histogram is used to collect numbers into a particular range of number. The histogram is defined as the chart that is used to represent the continuous data.
e.g. Consider the case of the data given below in the table that shows the data obtained in a class test of 35 students as:
The histogram drawn for this case is drawn as:
### Line Graph
A line graph is also known as a line chart. It is used to visualize the value of something over time. The line graph has a horizontal x-axis and a vertical y-axis. The point where axes intersect is called origin i.e. (0,0). Each axis having its own data type. For e.g. x-axis could have months, days, weeks and the y-axis may have a growth increase in shares and revenue.
All data value is represented in points and later they connected by line from one to other i.e. in “dot-to-dot” fashion.
e.g.
## Cartesian Planes and Coordinate Axes
### Cartesian Planes
A Cartesian plane is defined by the two perpendicular lines i.e. the x-axis(horizontal) and y-axis(vertical). With the help of these axes we can mark any point in the Cartesian plane.
The Cartesian plane is infinite however to shows this in book they put arrow in the end of the line. The Cartesian plane is divided into four quadrants.
### Coordinate Axes
It is x and y-axis if we label x and y-axis as in above diagram this framework is known as coordinate axis and the point of intersection of coordinate axes is known as origin.
### Representation of a point on the Cartesian Plane
In order to plot or represent a point on the cartesian plane first, the point must be in the form of (x,y) where the value of x is the coordinate of the given point on the x-axis while the value of y is the coordinate of the given point on the y-axis.
Thus, it is noted that the coordinates x and y helps to understand how far the given point is away from the origin with respect to both x and y-axes respectively.
e.g. Plot any point like (2,3) on the cartesian plane. For that consider 2 on x-axis and 3 on y-axis and plot as follows:
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# How do you write the product as a trinomial (3x + 2)(2x - 3)?
Jul 29, 2015
Multiply each term in one expression times each term in the other, then simplify by combining like terms.
#### Explanation:
There are other possible descriptions. Here is one of them:
To write the product $\left(3 x + 2\right) \left(2 x - 3\right)$ in another form, we need to multiply each terms in $3 x + 2$ times each term in $2 x - 3$.
(In algebra, 'terms' are things that are added together.)
So we need to multiply $3 x$ times $2 x$ and times $- 3$ (don't forget the minus sign!)
he we will multiply $2$ times $2 x$ and times $- 3$ (again with the minus sign)
So here's how we can write that:
$\left(3 x + 2\right) \left(2 x - 3\right) = 3 x \left(2 x\right) + 3 x \left(- 3\right) + 2 \left(2 x\right) + 2 \left(- 3\right)$
$= 6 {x}^{2} - 9 x + 4 x - 6$
Now I see two terms involving $x$, so we'll combine them into a single term:
$= 6 {x}^{2} - 5 x - 6$
Notice, now that we're finished, that the answer turns out to be a trinomial.
The instructions included the word 'trinomial' to try to make it clear that they didn't want us to rewrite the product in some other way -- like by changing the order, or by not combining like terms or doing some other way of rewriting. |
###### Alissa Fong
MA, Stanford University
Teaching in the San Francisco Bay Area
Alissa is currently a teacher in the San Francisco Bay Area and Brightstorm users love her clear, concise explanations of tough concepts
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Solving Two-step Equations - Concept
# Solving Single-step Equations - Problem 4
Alissa Fong
###### Alissa Fong
MA, Stanford University
Teaching in the San Francisco Bay Area
Alissa is currently a teacher in the San Francisco Bay Area and Brightstorm users love her clear, concise explanations of tough concepts
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Solving an equation and finding the value of the variable requires "undoing" what has been done to the variable. We do this by using inverse operations to isolate the variable. Remember that an equation is two expressions that are equal to each other. This means that when using inverse operations to isolate the variable, what is done to one side of the equation has to be done to the other side as well so that the equation stays balanced. In an equation, if "x" is in the denominator, start by multiplying both sides by x. This way, the "x" gets cancelled out from the denominator. Now you can solve the rest of the equation by continuing to use inverse operations. Make sure to keep in mind that any operation you do to one side of the equation, you must do to the other side as well. After solving for the variable, check your answer by plugging the value into the original equation. If the equation is true -- meaning the left side of the equation equals the right side of the equation -- then the solution is correct.
One of the more difficult solving equations problems that you'll see in single step equations is when x is in the denominator of the fraction and it's weird because not only do you have a fraction but x is in the bottom. But don't freak out, what this means is that 90 divided by some number gives you the answer 12. And the way you can get that x by itself is by undoing what's happening to it now.
So right now 90 is dividing x, I'm going to do the opposite which is multiplying both sides by x/1. So now those Xs cancel out and I'm left with 12x equals 90. That's an equation that I'm a lot more comfortable with because now there's no fraction going on, there's no division. To get x all by itself I need to undo that 12 times x business so now I have x equals 90/12 and depending on your textbook and your teacher that might be the appropriate final answer. A lot of textbooks say use your calculator to simplify or reduce all fractions. Since I'm a Math teacher I would make my students reduce that fraction. I want to show you how to do that.
First thing you do is in your brain you think about what number multiplies into 12 and into 90 and there's a couple of right answers. The first one that came to my brain was 3, so I'm going to divide both sides by 3 and I'll have, excuse me, divide top and bottom by three. That will give me 30/4 which again is not the most simple answer because there's still another number that goes into 30 and 4. That number is 2 so I need to divide by 2/2 and I'll get 15/2. That's my fractional answer that's most simplified. You might grab a calculator and turn that into 7.5. It's totally up to you and your teacher what form you leave it in.
Before we move on please make sure you check your solution. So go back to the original problem where we had, let's see 12 equals 90 over 15/2. That's ugly but don't freak out when you're dividing fractions that's the same thing as multiplying by the reciprocal right? So on this side I could simplify by doing 90 times the reciprocal. 12 is going to be equal to 90 times the reciprocal, so that's 2 on top of 15. I'm hoping 12 is equal to 180 divided by 15 and then if I'm right I'll be a happy camper. Let's just double check 180 divided by 15 you do indeed get 12. So that's how I know that I got 15/2 as the correct answer.
So the key to this problem, this type of problem where you have x in the denominator, is to do the opposite which is multiplying both sides by x. |
0
# A pre-algebra coin word problem
A girl saved nickels, dimes and quarters in a jar. She had as many quarters as dimes, but twice as many nickles as dimes. If the jar has 844 coins, how much money had she saved?
I am not sure how to put this into an equation or how to explain it. I already know that the quarters = 25 cents, etc.
### 1 Answer by Expert Tutors
Quang H. | Math and Science TutorMath and Science Tutor
4.0 4.0 (2 lesson ratings) (2)
0
First, we know that she has 844 coins in all, we know that she has the same amount of quarters as she does dimes, and she has twice the amount of nickels as she has dimes. The value of the coins does not matter right now, that comes later. We can start with this general equation:
n + d + q = 844
where the letters stand for what they obviously stand for.
Now, we can start to equate some things. We know that she has the same amount of quarters and dimes. Thus, we can produce the equality:
q=d=x
and I have added x here just to make the math clearer. X, here, will represent the number of both quarters and dimes.
Moving on, we know that she has twice as many nickels as dimes, and since the number of dimes is equal to x, then:
n = 2x
Now we can substitute the original terms in our first equation:
n + d + q = 844
would turn into:
2x + x + x = 844
We can then solve for x like this:
2x + x + x = 844
4x = 844
x = 211
Since x equals the number of dimes, then she has 211 dimes. And since she has as many dimes as quarters, then she has 211 quarters as well. And since she has twice as many nickels as she does dimes, the she has 221 * 2 = 422 nickels. If you want to check yourself, then you can add the number of coins up and see that 422 + 211 + 211 = 844 coins.
Now we can find the amount of money she has. We can multiply the value of the coins by the number of coins:
Dimes: 211 * \$0.10 = \$21.10
Quarters: 211 * \$0.25 = \$52.75
Nickels: 422 * \$0.05 = \$21.10
Adding all of those values up, we get \$94.95 |
# Free step by step math solver
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In this blog post, we discuss how Free step by step math solver can help students learn Algebra. There are many ways to solve problems involving interval notation. One popular method is to use a graphing calculator. Many graphing calculators have a built-in function that allows you to input an equation and then see the solution in interval notation. Another method is to use a table of values. This involves solving the equation for a few different values and then graphing the results. If the graph is a straight line, then the solution is simple to find. However, if the graph is not a straight line, then the solution may be more complicated. In either case, it is always important to check your work to make sure that the answer is correct.
Solving for a side in a right triangle can be done using the Pythagorean theorem. This theorem states that in a right triangle, the square of the length of the hypotenuse is equal to the sum of the squares of the lengths of the other two sides. This theorem can be represented by the equation: a^2+b^2=c^2. In this equation, c is the hypotenuse and a and b are the other two sides. To solve for a side, one would rearrange this equation to isolate the desired variable. For example, to solve for c, one would rearrange the equation to get c^2=a^2+b^2. To solve for a, one would rearrange the equation to get a^2=c^2-b^2. Once the equation is rearranged, one can then use basic algebraic techniques to solve for the desired variable. In this way, the Pythagorean theorem can be used to solve for any side in a right triangle.
A parabola solver is a mathematical tool that can be used to find the roots of a quadratic equation. Quadratic equations are equations that have the form ax^2 + bx + c = 0, where a, b, and c are constants. The roots of a quadratic equation are the values of x that make the equation equal to zero. A parabola solver can be used to find these roots by inputting the values of a, b, and c into the tool. The parabola solver will then output the roots of the equation. Parabola solvers can be found online or in mathematical textbooks.
In some cases, you may need to do a bit of research to find the answer. However, if you take your time and carefully read the question, you should be able to find the correct answer. With a little practice, you will be able to confidently answer math questions and improve your understanding of the subject.
Doing math homework can be a challenging and daunting task for many students. However, there are some simple tips that can make the process easier and more enjoyable. First, it is important to create a comfortable and well-lit workspace. This will help to reduce distractions and make it easier to focus on the task at hand. Second, it is helpful to break the homework down into smaller tasks. This will make the assignment seem less overwhelming and make it easier to track progress. Finally, it is important to ask for help when needed. Many students feel like they have to do everything on their own, but this is not the case. Asking for help from a teacher or tutor can be immensely helpful in understanding the material. By following these simple tips, Doing math homework can be a much more manageable task.
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# Percent - Grade 6 Math Questions and Problems With Solutions and Explanations
Detailed solutions and full explanations to grade 6 percent questions are presented.
30% of 30 = Solution 30% of 30 is equal to 30% × 30 = (30 / 100) × 30 = 900 / 100 = 9 150% of 60 = Solution 150% of 60 is equal to 150% × 60 = (150 / 100) × 60 = (150 × 60) / 100 = 90 1/4 = 4% 1% 0.25% 25% Solution We need to change the fraction 1 / 4 into a fraction with denominator 100 which is a percent 1 / 4 = (1 × 25) / (4 × 25) = 25 / 100 = 25% 0.05 = 50% 500% 5% 0.5% Solution We need to change the decimal number 0.05 into a fraction with denominator 100 which is a percent 0.05 = 0.05 / 1 = (0.05 × 100) / (1 × 100) = 5 / 100 = 5% If 100% of a number is 15, what is 50% of the number? Solution Let n be the number 100% of a number is 15 is written as: 100% × n = 15 100% = 100 / 100 = 1 Hence 100% × n = n = 15 We now need to find 50% of the number 50% of 15 = (50 / 100) × 15 = 750 / 100 = 7.5 NOTE that the above problem could be solved by noting that 50% of something is half of 100% of the same thing. Hence if 100% of something is 15 then 50% of the same thing is haflf of 15 / 2 = 7.5 If 10% of a number is 7, what is 80% of the number? Solution Note that 80% of something is 8 times 10% of the same thing. Hence if 10% of a number is 7 then 80% of the same number is given by 8 × 7 = 56 Which is the greatest? 90% of 10 6% of 1000 5% of 1400 3% of 2500 Solution Express as fractions with same denominator 100 and then compare 90% of 10 = (90 / 100) × 10 = 900 / 100 6% of 1000 = (6 / 100) × 1000 = 6000 / 100 5% of 1400 = (5 /100) × 1400 = 7000 / 100 3% of 2500 = (3 / 100) × 2500 = 7500 / 100 The largest fraction is 7500 / 100 which correspond to 3% of 2500. Hence 3% of 2500 is the largest. The original price of a toy was $15. If the price is reduced by 20%, what is the new price of the toy? Solution 20% of price is 20% × 15 =$3 The original price was $15 and was reduced by 20% or$3. Hence the new price after reduction is 15 - 3 = $12 George bought a car at$5000 and sold it at $5500. What benefit, in percent, did he make? Solution The benefit in dollars is$5500 - $5000 =$500 Express $500 as a percent of$5000 as follows 500 / 5000 = 5 / 50 = 10 / 100 = 10% If 20% of n is equal to 40, what is n? Solution "20% of n is equal to 40" is written as 20% × n = 40 (20 / 100) × n = 40 Rewrite the above with fraction on the right side with denominator equal to 100 20 n / 100 = 40 / 1 = 4000 / 100 Since the two fractions have same denominator and are equal, their numerator must be equal. Hence 20 n = 4000 which means that n = 200. 200 2000 800 80 The price of a T-shirt was $20. It was first increased by 20%. They did not sell well the shop owner decreased the price by 20%. What is the new price of the T-shirt?$20 $22$21 \$19.20 What percent of 1 hour is 15 minutes? 50% 15% 75% 25%
1. C
2. D
3. D
4. C
5. A
6. B
7. D
8. A
9. B
10. A
11. D
12. D
More Middle School Math (Grades 6, 7, 8, 9) - Free Questions and Problems With Answers
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NCERT Solutions Maths Class 10 Exercise 13.2
# NCERT Solutions Maths Class 10 Exercise 13.2
## Maths Class 10 Exercise 13.2
1. A solid is in the shape of a cone standing on a hemisphere with both their radii being equal to 1 cm and the height of the cone is equal to its radius. Find the volume of the solid in terms of π.
Solution: Radius of the hemisphere (r) = 1 cm
Volume of the hemisphere = 2/3 × Ï€r3
= 2/3 × Ï€(1)3
= 2Ï€/3 cm3
Radius of the base of the cone (r) = 1 cm
Height of the cone = 1 cm
Volume of the cone = 1/3 × Ï€r2h
= 1/3 × Ï€(1)2 × 1
= π/3 cm3
Volume of the solid = Volume of the hemisphere + Volume of the cone
= 2π/3 + π/3
= π cm3
2. Rachel, an engineering student, was asked to make a model shaped like a cylinder with two cones attached at its two ends by using a thin aluminium sheet. The diameter of the model is 3 cm and its length is 12 cm. If each cone has a height of 2 cm, find the volume of air contained in the model that Rachel made. (Assume the outer and inner dimensions of the model to be nearly the same.)
Solution: For upper and lower conical portions, radius of the base (r) = 1.5 cm
Height of the cones (h1) = 2 cm
Volume of both the cones = 2 × 1/3 × Ï€r2h1
= 2/3 × Ï€(1.5)2 × 2
= 3Ï€ cm3
Radius of the base of the cylinder (r) = 1.5 cm
Height of the cylinder (h2) = 12 – (2 + 2) = 8 cm
Volume of the cylinder = Ï€r2h2 = Ï€(1.5)2 × 8 = 18Ï€ cm3
Volume of the model = Volume of both the cones + Volume of cylinder = 3Ï€ + 18Ï€
= 21Ï€ cm3
= 21 × 22/7
= 66 cm3
3. A gulab jamun, contains sugar syrup up to about 30% of its volume. Find approximately how much syrup would be found in 45 gulab jamuns, each shaped like a cylinder with two hemispherical ends, with length 5 cm and diameter 2.8 cm (see figure).
Solution: Volume of a gulab jamun = Volume of two hemispheres + Volume of a cylinder
= 2 × 2/3 × Ï€r3 + Ï€r2h
= 4/3 × Ï€(1.4)3 + Ï€(1.4)2 × 2.2
= Ï€(1.4)2[4 × 1.4/3 + 2.2]
= 1.96Ï€(5.6 + 6.6)/3
= 1.96Ï€(12.2)/3 cm3
Volume of 45 gulab jamuns = 45 × 1.96Ï€(12.2)/3 cm3
= 15 × 1.96Ï€(12.2) cm3
= 15 × 1.96 × 12.2 × 22/7 cm3
= 1127.28 cm3
Volume of the syrup in 45 gulab jamuns = 1127.28 × 30/100
= 338.184 cm3
= 338 cm3 (approx.)
4. A pen stand made of wood is in the shape of a cuboid with four conical depressions to hold pens. The dimensions of the cuboid are 15 cm by 10 cm by 3.5 cm. The radius of each of the depressions is 0.5 cm and the depth is 1.4 cm. Find the volume of wood in the entire stand (see figure).
Solution: The dimensions of the cuboid = 15 cm by 10 cm by 3.5 cm
Volume of the cuboid = l × b × h
= 15 × 10 × 3.5
= 525 cm3
Volume of the conical depression = 1/3 × Ï€r2h
= 1/3 × 22/7 × 0.5 × 0.5 × 1.4
= 11/30 cm3
Volume of the four conical depressions = 4 × 11/30 = 1.47 cm3
Volume of the wood in the entire stand = 525 – 1.47 = 523.53 cm3
5. A vessel is in the form of inverted cone. Its height is 8 cm and the radius of the top, which is open, is 5 cm. It is filled with water up to the brim. When lead shots, each of which is a sphere of radius 0.5 cm are dropped into the vessel, one-fourth of the water flows out. Find the number of lead shots dropped in the vessel.
Solution: Radius of the top of the conical vessel (r) = 5 cm and its height (h) = 8 cm
Volume of the conical vessel = 1/3 × Ï€r2h
= 1/3 × Ï€(5)2 × 8
= 200Ï€/3 cm3
Volume of spherical lead shot = 4/3 × Ï€R3
= 4/3 × Ï€(0.5)3
= π/6 cm3
Volume of water that flows out = ¼ × Volume of the cone
= ¼ × 200Ï€/3 cm3
= 50Ï€/3 cm3
Let the number of lead shots dropped in the vessel be n.
Therefore, n × Ï€/6 = 50Ï€/3
n = 50Ï€/3 × 6/Ï€
n = 100
Hence, 100 lead shots dropped in the vessel.
6. A solid iron pole consists of a cylinder of height 220 cm and base diameter 24 cm, which is surmounted by another cylinder of height 60 cm and radius 8 cm. Find the mass of the pole, given that 1 cm3 of iron has approximately 8 g mass. (Use π = 3.14)
Solution: Base radius of lower cylinder (r) = 24/2 = 12 cm
Height of the lower cylinder (h) = 220 cm
Volume of the lower cylinder = πr2h
= Ï€(12)2 × 220
= 31680Ï€ cm3
Base radius of upper cylinder (R) = 8 cm
Height of upper cylinder (H) = 60 cm
Volume of upper cylinder = πR2H
= Ï€(8)2 × 60
= 3840Ï€ cm3
Volume of the solid iron pole = Volume of lower cylinder + Volume of upper cylinder
= 31680Ï€ + 3840Ï€
= 35520Ï€
= 35520 × 3.14
= 111532.8 cm3
Since the mass of 1 cm3 of iron = 8 g
Therefore, the mass of 111532.8 cm3 of iron = 111532.8 × 8 g
= 892262.4 g
= 892.26 kg
Hence, the mass of the pole is 892.26 kg.
7. A solid consisting of a right circular cone of height 120 cm and radius 60 cm standing on a hemisphere of radius 60 cm is placed upright in a right circular cylinder full of water such that it touches the bottom. Find the volume of water left in the cylinder, if the radius of the cylinder is 60 cm and its height is 180 cm.
Solution: Radius of the base of the right circular cone (r) = 60 cm
And height of the right circular cone (h1) = 120 cm
Volume of the right circular cone = 1/3 × Ï€r2h1
= 1/3 × Ï€(60)2 × 120
= 144000Ï€ cm3
Radius of the base of the hemisphere (r) = 60 cm
Volume of the hemisphere = 2/3 × Ï€r3
= 2/3 × Ï€(60)3
= 144000Ï€ cm3
Radius of the base of the right circular cylinder (r) = 60 cm
And height of the right circular cylinder (h2) = 180 cm
Volume of the right circular cylinder = πr2h2
= Ï€(60)2 × 180
= 648000Ï€ cm3
Now, volume of water left in the cylinder = Volume of right circular cylinder – (Volume of right circular cone + Volume of hemisphere)
= 648000Ï€ – (144000Ï€ + 144000Ï€)
= 648000Ï€ – 288000Ï€
= 360000Ï€ cm3
= 0.36Ï€ m3
= 0.36 × 22/7 m3
= 1.131 m3 (approx.)
8. A spherical glass vessel has a cylindrical neck 8 cm long, 2 cm in diameter; the diameter of the spherical part is 8.5 cm. By measuring the amount of water it holds, a child finds its volume to be 345 cm3. Check whether she is correct, taking the above as the inside measurements, and π = 3.14.
Solution:
The amount of water the vessel holds = Volume of spherical part + Volume of cylindrical part
= (4/3) × Ï€r3 + Ï€r2h
= (4/3) × Ï€(8.5/2)3 + Ï€(1)2 × 8
= 4/3 × 3.14 × 4.25 × 4.25 × 4.25 + 3.14 × 8
= 321.39 + 25.12
= 346.51 cm3
Hence, the child is not correct. The correct volume is 346.51 cm3. |
## Thursday, July 2, 2009
### Vedic Mathematics Lesson 3: Multiplication Part 1
You can read about my interest in Vedic Mathematics in this earlier post, and read Lesson 1 and Lesson 2 to appreciate the power of Vedic Mathematics, and to learn the basics that we are going to use in this lesson.
In this lesson, our main goal will be to learn to multiply large numbers that are just under a power of 10 without doing any long multiplication. Imagine trying to compute 9986 x 9997 without going through 4 sets of laborious multiplications and then a laborious addition. When we are done with this lesson, we will be able to do the above problem in our minds and find the answer in about 3 to 5 seconds.
As always, it is best to start from the basics and build up to what we want to accomplish. Let us go through a simple example to illustrate the method, then I will explain the algebraic logic behind why the method works, then we will apply it to bigger problems and also explore some additional considerations.
Assume that we have to find the answer to 9x8. This is quite simple to do for most people who know their multiplication tables and should not require any Vedic Mathematics to accomplish. However, we will use this problem to illustrate the method and extend it to products that would not normally be covered by multiplication tables.
Follow these steps to find 9x8:
• Find the appropriate base for our calculations. The base will be the power of 10 that is closest to the numbers to be multiplied. In this particular case, our base will be 10 itself.
• Put the numbers in two rows on the left hand side. In the middle column, put a "-" if the number is less than the base and a "+" if the number is more than the base. In our case, the middle column for both rows will be "-" since they are both less than 10.
• If the middle column is a "-", in the right hand side, write the 10's complement of the number, or the deficit from the base.
• If the middle column is a "+", in the right hand side, write the amount by which the number is greater than the base (we will deal with this case in a subsequent lesson).
After these 4 steps, we get the figure below:
9 - 1
8 - 2
Our first number is 9. It is less than our base, 10, so there is a "-" in the middle column. In the right hand side, we have the 10's complement of 9, which is 1. Similarly, in the second row, we have our second number, 8, a "-" because 8 is smaller than our base, 10, and the 10's complement of 8, which is 2.
Now, the product will have two parts, a left hand part and a right hand part. We can draw a vertical line on the third row to demarcate the two parts (this will not be necessary as we become better at the method with practice and learn to do the whole thing mentally). So, now we have the figure below:
9 - 1
8 - 2
-------
|
The left hand side of the answer can be found in one of 4 different ways:
• Add the numbers on the left hand side and subtract the base from the answer. This gives us 9 + 8 - 10 = 7.
• Add the deficiences on the right hand side and subtract that from 10. This gives us 10 - 2 - 1 = 7.
• Cross subtract the deficiency on the first line from the second number. This gives us 8 - 1 = 7.
• Cross subtract the deficiency on the second line from the first number. This gives us 9 - 2 = 7.
So, our figure now looks as below:
9 - 1
8 - 2
--------
7 |
The right hand side of the answer can be found with one simple calculation:
• Simply multiply the deficits by each other to get the right hand side of the answer.
Our figure will now look as below:
9 - 1
8 - 2
--------
7 |2
The answer to the problem is the combination of the left and right hand sides of the answer above, which gives us 72. This is obviously easy to verify as the correct answer not only with a calculator, but also by using our memory of multiplication tables.
What is the algebra behind why this method works? Assume that our base is b. Let us assume that the two numbers whose product we have to find are y and z. Let us assume that d and e are the deficits of y and z from our base, b. So, we are trying to find the product y x z which is the same as (b - d) x (b - e). We see that this can be written as b x (b - d - e) + d x e.
Which is precisely what we did using the method illustrated in the steps above: We found the left hand side as the difference between the base and the sum of the deficits, then multiplied it by the base so that it becomes the left hand side of the answer. Then we multiplied the deficits by each other and made it the right hand side of the answer.
Now, let us illustrate the method on more complex problems. These problems will not only give us practice, but also illustrate how to deal with some issues we might encounter along the way.
Let us take 96 x 95. First we choose 100 as the base. Then we write the problem out as below:
96 - 4
95 - 5
--------
|
Note that both numbers are below the base, hence the "-" sign in the middle column. 4 and 5 are the 10's complements of 96 and 95 respectively (their deficits from our base of 100).
Now, we find the left hand side of the answer using any one of the 4 methods outlined earlier. We get either 96 + 95 - 100 = 91, or 95 - 4 = 91, or 96 - 5 = 91 or 100 - 4 - 5 = 91. Our figure becomes:
96 - 4
95 - 5
---------
91 |
Now we multiply the deficits by each other to find the right hand side of our answer. It is 4 x 5 = 20. So, our final answer is 9120, and our figure looks as below:
96 - 4
95 - 5
---------
91 | 20
Now, let us apply this method to problems which present special cases we have not encountered so far.
First, let us try to do 99 x 99. Following the methodology outlined above, we get the figure below:
99 - 1
99 - 1
--------
98 | 1
We quickly realize that 981 is not the answer to the problem. Is there something wrong with the Vedic method? Not really. Looking at the algebraic explanation of the method, we find that the answer is actually the left hand side of the answer multiplied by the base, added to the right hand side of the answer. This means that the right hand side of the answer has to have exactly as many digits as the number of zeroes in the base (when we multiply the left hand side by the base, in this case, 100, we get 9800. The right hand side is actually added to 9800, not just appended to 98 though that is the practical effect when the right hand side contains exactly two digits). In this case, the base is 100, so the right hand side has to have 2 digits. We accomplish this by padding the answer with zeroes to the left until we get the requisite number of digits. Thus our figure becomes:
99 - 1
99 - 1
---------
98 | 01
Now our answer is 9801 and this can be verified to be correct using a calculator, or by long multiplication.
Next, let us see what happens when we try to multiply 90 x 88. We get the figure below:
90 - 10
88 - 12
---------
78 | 120
Obviously 78120 is not the correct answer to the problem. But the algebraic explanation of the problem comes to the rescue again. We see that the answer, algebraically, is actually 78 x 100 + 120. The right hand side has to be restricted to two digits. In our case, we have 3 digits, so retain the right most 2 digits and use any leftover digits as carryover to increase the left hand side by. This leads, in this case, to the right hand side becoming 20 and the 1 becoming a carryover digit. Adding 1 to the left hand side gives us 79. Our figure now becomes:
90 - 10
88 - 12
----------
78+1 | 20
The answer is 7920, which can be verified to be correct.
Now, let us tackle the problem we initially posed in this lesson: what is 9986 x 9997? We quickly realize that the base required for this problem is 10000. Calculating the deficits of the numbers from 10000, we draw the figure below:
9986 - 14
9997 - 3
--------------
|
Next we find the left and right hand sides of the answer as below:
9986 - 14
9997 - 3
--------------
9983 | 42
Now we note that the right hand side contains only 2 digits whereas 10000 has 4 zeroes. So, we need to pad 42 out to 4 digits, giving us 0042. The figure now looks as below:
9986 - 14
9997 - 3
--------------
9983 | 0042
This automatically leads to the answer 99830042 which can be verified using a calculator.
In the interest of a little more practice, and the application of another of the special cases we are likely to encounter, let us try to find 9900 x 9900. We use 10000 as the base, which leads to the figure below:
9900 - 100
9900 - 100
-------------
9800 | 10000
Immediately, we see that the right hand side actually has 5 digits, one more than it should have. Using the rule regarding carryovers, we modify the diagram as below:
9900 - 100
9900 - 100
--------------
9800 + 1 | 0000
This leads to the answer 98010000 which can be verified to be correct.
Now, let us do 9000 x 9000. Again the base is 10000, which leads to the figure below:
9000 - 1000
9000 - 1000
----------------
8000 | 1000000
Our right hand side has 3 extra digits. Moving them to the left hand side using the rule regarding carryovers, we get:
9000 - 1000
9000 - 1000
----------------
8000 + 100|0000
This gives us the correct answer of 81000000.
In subsequent lessons, we will expand on what we learned in this lesson so that we can handle a wider variety of problems. After all, not all the numbers we need to multiply are just under a power of 10! Practice makes perfect, so happy practicing and good luck!!
Dhwani Kapoor said...
the simple language and examples are good for a new learner.
thanks!
Blogannath said...
Thank you. I am glad you found the post simple and helpful.
Anonymous said...
it will be helpful if you post the problems over here....or suggest me a good book to do the same...where i can apply some of these tricks very easily...
Blogannath said...
I have provided only the method along with some examples in my blog, not actual exercises with sample problems. If you are interested in sample problems, you can try the series of books called "Vedic Mathematics For Schools" by James T. Glover. One of them is available on Amazon.com, and the whole series may be available in your local bookstore. Good luck.
Anonymous said...
Nice post...
How about doing multiplication of 11x88, using base as 100..??
Blogannath said...
As the second paragraph of my post explains, this post is "to learn to multiply large numbers that are just under a power of 10". As such, 11 and 88 are not two numbers that are just below a power of 10, so you have to read other posts in this series to figure out how to do that without long multiplication. This post can not help you with that.
Ketul Patel said...
very detailed explanation. Thank you very much. if you consider twisting this info in a manner more helpful for those preparing for competitive exams, it would be even better. The present one is excellent, but this is just a suggestion.
Anonymous said...
Superb blog !
venkatesh said...
have u mentioned anythng regarding multiplication of numbers with different base... for example 125*525? pls help me...
Shanthi said...
Great Article thanks a lot for sharing with us
William Wheeler said...
You have taken a great deal of trouble to explain the methods very well. Couldn't ask for more! Very well done. And thank you very much.
## Content From TheFreeDictionary.com
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Match each word in the left column with its synonym on the right. When finished, click Answer to see the results. Good luck!
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Which of the following numbers are not perfect cubes?
Question:
Which of the following numbers are not perfect cubes?
(i) 64
(ii) 216
(iii) 243
(iv) 1728
Solution:
(i)
On factorising 64 into prime factors, we get:
$64=2 \times 2 \times 2 \times 2 \times 2 \times 2$
On grouping the factors in triples of equal factors, we get:
$64=\{2 \times 2 \times 2\} \times\{2 \times 2 \times 2\}$
It is evident that the prime factors of 64 can be grouped into triples of equal factors and no factor is left over. Therefore, 64 is a perfect cube.
(ii)
On factorising 216 into prime factors, we get:
$216=2 \times 2 \times 2 \times 3 \times 3 \times 3$
On grouping the factors in triples of equal factors, we get:
$216=\{2 \times 2 \times 2\} \times\{3 \times 3 \times 3\}$
It is evident that the prime factors of 216 can be grouped into triples of equal factors and no factor is left over. Therefore, 216 is a perfect cube.
(iii)
On factorising 243 into prime factors, we get:
$243=3 \times 3 \times 3 \times 3 \times 3$
On grouping the factors in triples of equal factors, we get:
$243=\{3 \times 3 \times 3\} \times 3 \times 3$
It is evident that the prime factors of 243 cannot be grouped into triples of equal factors such that no factor is left over. Therefore, 243 is a not perfect cube
(iv)
On factorising 1728 into prime factors, we get
$1728=2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 3 \times 3 \times 3$
On grouping the factors in triples of equal factors, we get:
$1728=\{2 \times 2 \times 2\} \times\{2 \times 2 \times 2\} \times\{3 \times 3 \times 3\}$
It is evident that the prime factors of 1728 can be grouped into triples of equal factors and no factor is left over. Therefore, 1728 is a perfect cube.
Thus, (iii) 243 is the required number, which is not a perfect cube. |
## Knowing Our Numbers
NCERT Ex 1.2 Solutions
CBSE Class 6 Maths
Q1. A book exhibition was held for four days in a school. The number of tickets sold at the counter on the first, second, third and final day was respectively 1094, 1812, 2050 and 2751. Find the total number of tickets sold on all the four days.
Number of tickets sold on first day = 1,094
Number of tickets sold on second day = 1,812
Number of tickets sold on third day = 2,050
Number of tickets sold on fourth day = 2,751
────────
Total tickets sold = 7,707
────────
∴ 7,707 tickets were sold on all the four days.
Q2: Shekhar is a famous cricket player. He has so far scored 6980 runs in test matches. He wishes to complete 10,000 runs. How many more runs does he need?
Runs to achieve = 10,000
Runs scored = – 6,980
────────
Runs required = 3,020
────────
∴ he needs 3,020 more runs.
Q3: In an election, the successful candidate registered 5,77,500 votes and his nearest rival secured 3,48,700 votes. By what margin did the successful candidate win the election?
Number of votes secured by successful candidates = 5,77,500
Number of votes secured by his nearest rival = – 3,48,700
────────
Margin between them = 2,28,800
────────
Thus, the successful candidate won by a margin of 2,28,800 votes.
Q4: Kirti bookstore sold books worth Rs 2,85,891 in the first week of June and books worth Rs 4,00,768 in the second week of the month. How much was the sale for the two weeks together? In which week was the sale greater and by how much?
Books sold in first week = 2,85,891
Books sold in second week = + 4,00,768
────────
Total books sold = 6,86,659
────────
Since, 4,00,768,> 2,85,891, ∴ sale of second week is greater than that of first week.
Books sold in second week = 4,00,768
Books sold in first week = – 2,85,891
────────
More books sold in second week = 1,14,877
────────
∴ 1,14,877 more books were sold in second week
Q5: Find the difference between the greatest and the least number that can be written using the digits 6, 2, 7, 4, 3 each only once.
Greatest five-digit number using digits 6,2,7,4,3 = 76432
Smallest five-digit number using digits 6,2,7,4,3 = – 23467
────────
Difference = 52965
────────
Thus the difference is 52965.
Q6: A machine, on an average, manufactures 2,825 screws a day. How many screws did it produce in the month of January 2006?
Number of screws manufactured in one day = 2,825
Number of days in the month of January (31 days) = 2,825 × 31
= 87,575
∴ the machine produced 87,575 screws in the month of January.
Q7: A merchant had Rs 78,592 with her. She placed an order for purchasing 40 radio sets at Rs 1200 each. How much money will remain with her after the purchase?
Cost of one radio = ₹ 1200
Cost of 40 radios = 1200 x 40 = ₹ 48,000
Total money with merchant = ₹ 78,592
Money spent by her = – ₹ 48,000
────────
Money left with her = ₹ 30,592
────────
Thus ₹ 30,592 will remain with her after the purchase.
Q8: A student multiplied 7236 by 65 instead of multiplying by 56. By how much was his answer greater than the correct answer? (Hint: Do you need to do both the multiplications?)
Wrong answer = 7236 × 65 Correct answer = 7236 × 56
7236 7236
× 65 × 56
──────── ────────
36180 43416
43416x 36180x
──────── ────────
470340 405216
──────── ────────
Difference in answers = 470340 – 405216
= 65,124
Q9: To stitch a shirt, 2 m 15 cm cloth is needed. Out of 40 m cloth, how many shirts can be
stitched and how much cloth will remain?
(Hint: convert data in cm.)
Cloth required to stitch one shirt = 2 m 15 cm
= 2 × 100 cm + 15 cm
= 215 cm
Length of cloth = 40 m = 40 x 100 cm = 4000 cm
Number of shirts can be stitched = 4000 ÷ 215
18
────────
215)4000
215
────
1850
1720
──────
130
Thus 18 shirts can be stitched and 130 cm (1 m 30 cm) cloth will remain.
Q10: Medicine is packed in boxes, each weighing 4 kg 500g. How many such boxes can be loaded in a van which cannot carry beyond 800 kg?
The weight of one box = 4 kg 500 g = 4 x 1000 g + 500 g = 4500 g
Maximum load can be loaded in van = 800 kg = 800 x 1000 g = 800000 g
Number of boxes = 800000 ÷ 4500
177
────────
4500)800000
4500
─────
35000
31500
──────
35000
31500
────────
3500
Thus 177 boxes can be loaded.
Q11: The distance between the school and the house of a student’s house is 1 km 875 m. Everyday she walks both ways. Find the total distance covered by her in six days.
Distance between school and home = 1.875 km
Distance between home and school = + 1.875 km
Total distance covered in one day = 3.750 km
Distance covered in six days = 3.750 × 6 = 22.500 km
Thus 22 km 500 m distance covered in six days.
Q12: A vessel has 4 litres and 500 ml of curd. In how many glasses, each of 25 ml capacity, can it be filled?
Capacity of curd in a vessel = 4 litres 500 ml = 4 x 1000 ml + 500 ml = 4500 ml
Capacity of one glass = 25 ml
Number of glasses can be filled = 4500 ÷ 25
180
────────
25)4500
25
─────
200
200
────
0
Thus 180 glasses can be filled by curd. |
Mechanics
Scientific notation and significant figures
Scientific notation and significant figures are two important terms in physics. In scientific notation, numbers are expressed by some power of ten multiplied by a number between 1 and 10, while significant figures are accurately known digits and first doubtful digit in any measurement.
Scientific notation definition
In scientific notation, a number is expressed as some power of ten multiplied by a number between 1 and 10.
A simple and scientific method to write small or large numbers is to express them in some power of 10. The distance of the moon from Earth is 384000000 meters. This is a large number, it can also be expressed as 3.84×10 8 m. This way of expressing the number is scientific notation.
Why we use scientific notation?
We use as it saves writing down large numbers of zeroes.
Scientific notation examples
• The number 15000000 km can be written as 1.5 ×1011 m.
• 0.00000548 S can be written as: 5.48 ×10-6 s.
• Mass of Earth is written as: 6×10 24 kg.
• The radius of earth is written as:6.4 ×104 m.
Decimal to scientific notation
• 0.2
Solution: 0.2 = 2 × 10¹
• 0.006
Solution: 0.006=6 × 103
• 0.00063
Solution: 6 × 10 4
• 0.00000678
Solution: 0.00000678 = 6.78 ×10 8
Scientific notation to decimal
• 3 ×10-1
Solution:3 ×10-1 =0.3
• 6 × 10- 4
Solution: 0.0006
Scientific notation to standard notation
In standard notation there is only one non zero number on the left side of the zero.Below examples has been given:
• 1168 ×10 -27
Solution: 1168 ×10 -27 =1.168 ×10 -24
• 725 ×10 -5
Solution: 725 ×10 -5
=7.25 ×10 -3
What is significant figure?
A significant figure is one, which is known to be reasonable and reliable.
or
All the accurately known digits and first doubtful digit in an expression are called significant figures.
Physics is based on measurements. But unfortunately when a physical quantity is measured, then there is inevitably some uncertainty about its find value. This uncertainty may be due to the number of reasons.
One reason is the type of instrument is being used. We know that every measuring instrument is calibrated to a certain smallest division and this fact put a limit to the degree of accuracy which may be achieved while measuring with it. Suppose we want to measure the length of a straight line with the help of a rod calibrated in millimeters. Let the end point of the line lies between 10.3 and 10.4 cm marks. By convention, if the end of the line does not touch or cross the midpoint of the smallest division, the reading is confined to the previous division. In case the end of the line seems to be touching or have crossed the midpoint, the reading is extended to the next division.
By applying the above rule the position of the edge of a line recorded as 12.7 cm with the help of a meter rod calibrated in millimeters may lie between 12.65 cm and 12.75 cm. Thus in this example the maximum uncertainty ±0.05 cm. It is, in fact, equivalent to an uncertainty of 0.1 cm equal to the least count of the instrument divided into two parts, half above and half below the recorded reading.
The uncertainty or accuracy in the value of a measured quantity can be indicated conveniently by using significant figures. The recorded value of the length of the straight line I.e., 12.7 cm contains three digits (1,2,7) out of which two digits (1 and 2) are accurately known while the third digit (7) is a doubtful one. As a rule:
“In any measurement, the accurately known digits and the first doubtful digit are called significant figures.”
In other words, a significant figure is the one which is known to be reasonably reliable. If the above-mentioned measurement is taken by a better measuring instrument which is exact up to a hundredth of a centimeter, it would have been recorded as 12.70 cm. In this case, the number of significant figures is four. Thus, we can say that as we improve the quality of our measuring instrument and techniques, we extend the measured result to more and more significant figures and correspondingly improve the experimental accuracy of the result. While calculating a result from the measurements, it is important to give due attention to significant figures and we must know the following rules in deciding how many significant figures are to be retained in the final result.
Significant figures rules
• Digits other than zero are always significant.
• Zeroes between significant digits are also significant.
• Zero on the left of the significant figures is not significant.
• Zero on the right of the significant figure is not significant.
• Zero on the right of a fractional number is significant
Significant figures examples
Significant figures examples according to the rules which are mentioned above are:
• In 15.2 significant figures are 3 .
• In 203 significant figures are 3.
• In 2100 significant figures are 2
• In 21.00 significant figures are 4.
• In 0.002 significant figures are 1.
• In 10.40 significant figures are 4.
Related posts:
Vernier caliper least count formula Micrometer screw gauge least count formula types of physical quantities and their examples Difference between accuracy and precision Difference between screw gauge and vernier caliper |
Q:
# How do you factor an equation by grouping?
A:
To factor an equation by grouping, separate the polynomial expression into two binomials, and find the zeroes of the equation that cross the x-axis. Factoring an equation by grouping involves using the master product of the first and last terms.
## Keep Learning
1. Find the master product
The master product is the product of the coefficient of a and c, when the equation is in the standard form ax^2 + bx + c = 0. For example, the equation 3x^2 + 14x + 15 has a master product of 45.
2. Determine what factors of ac add up to b
The coefficient of b is 14. Factors of 45 that add to 14 are 9 and 5. Rewrite the term 14x as the sum of 5x and 9x. Now your equation should look like 3x^2 + 9x + 5x + 15.
3. Separate the term into binomials
Group the polynomial into binomials. This leaves you with (3x^2 + 9x) + (5x + 15). After doing this, use the greatest common factor and distributive property to simplify the terms further. This transforms the expression to 3x(x + 3) + 5(x + 3).
4. Group the factored terms into binomials
Because 3x and 5 are separated, they can be placed into their own binomial, leaving the factored expression as (3x + 5)(x + 3) = 0
5. Find the zeroes of the expression
This is the value of x that would make the expression equal zero. Either term of x works, because anything multiplied by zero is zero. The zeroes of the example are -1.6666 and -3.
Sources:
## Related Questions
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• A: In math, expanded form can refer to any type of expression, equation or notation that is completely broken down into its individual parts. Expanded form is... Full Answer >
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## What is the limit of the product of two functions?
The Product Law basically states that if you are taking the limit of the product of two functions then it is equal to the product of the limits of those two functions. [f(x) · g(x)] = L · M.
Is the limit of a product the product of the limits?
The multiplication rule for limits says that the product of the limits is the same as the limit of the product of two functions. That is, if the limit exists and is finite (not infinite) as x approaches a for f(x) and for g(x), then the limit as x approaches a for fg(x) is the product of the limits for f and g.
### How do you prove a limit is correct?
We prove the following limit law: If limx→af(x)=L and limx→ag(x)=M, then limx→a(f(x)+g(x))=L+M. Let ε>0. Choose δ1>0 so that if 0<|x−a|<δ1, then |f(x)−L|<ε/2….Proving Limit Laws.
Definition Opposite
1. For every ε>0, 1. There exists ε>0 so that
2. there exists a δ>0, so that 2. for every δ>0,
READ ALSO: What type of hair does Sakura have?
How do we find the product of two or more functions?
As you might guess, finding the product of functions is as simple multiplying the functions together. When you multiply two functions together, you’ll get a third function as the result, and that third function will be the product of the two original functions.
## How do you use limit properties?
How To: Given a function containing a polynomial, find its limit.
1. Use the properties of limits to break up the polynomial into individual terms.
2. Find the limits of the individual terms.
4. Alternatively, evaluate the function for a .
How do you do limits?
For example, follow the steps to find the limit:
1. Find the LCD of the fractions on the top.
2. Distribute the numerators on the top.
3. Add or subtract the numerators and then cancel terms.
4. Use the rules for fractions to simplify further.
5. Substitute the limit value into this function and simplify.
### How do you prove a function exists?
How to approach questions that ask to prove a function exists?
1. if r(y)=r(x)⇒h(y)=h(x)
2. h(y)=g(r(y))
3. Assume there exists a function g:Q→T . Then r(x)=r(y)⇒g(r(x))=g(r(y))
4. The above does not look helpful in proving the conclusion.
READ ALSO: Can we crack competitive exam without coaching?
What condition do we need to show in order to prove a limit?
In general, to prove a limit using the ε \varepsilon ε- δ \delta δ technique, we must find an expression for δ \delta δ and then show that the desired inequalities hold. The expression for δ \delta δ is most often in terms of ε , \varepsilon, ε, though sometimes it is also a constant or a more complicated expression.
## How do you solve a product function?
To multiply a function by another function, multiply their outputs. For example, if f (x) = 2x and g(x) = x + 1, then fg(3) = f (3)×g(3) = 6×4 = 24. fg(x) = 2x(x + 1) = 2×2 + x.
How do you know when to use the product rule?
The Product Rule must be utilized when the derivative of the product of two functions is to be taken. The Product Rule says that the derivative of a product of two functions is the first function times the derivative of the second function plus the second function times the derivative of the first function.
### Does the limit of a product of functions equal the product?
The limit of a product of functions equals the product of the limits: Is this proof rigorous? All the proofs I’ve seen so for the limit of a product of functions equaling the product of the limits are based on the following: | g ( x) | = | g ( x) − G + G | ≦ | g ( x) − G | + | G | < 1 + | G | when 0 < | x − x 0 | < δ 3. This settles the proof.
READ ALSO: Why do some houses not have mailboxes?
What is the product rule for derivatives and exponents?
The product rule for different functions such as derivatives, exponents, logarithmic functions are given below: For any two functions, say f (x) and g (x), the product rule is D [f (x) g (x)] = f (x) D [g (x)] + g (x) D [f (x)] If m and n are the natural numbers, then x n × x m = x n+m.
## How do you use the product rule in calculus?
In Calculus, the product rule is used to differentiate a function. When a given function is the product of two or more functions, the product rule is used. If the problems are a combination of any two or more functions, then their derivatives can be found using Product Rule. The derivative of a function h(x) will be denoted by D {h(x)} or h'(x).
What is the limit of the function as approaches 2?
Let’s first take a closer look at how the function behaves around in (Figure). As the values of approach 2 from either side of 2, the values of approach 4. Mathematically, we say that the limit of as approaches 2 is 4. Symbolically, we express this limit as |
# Linear Algebra 5 | Orthogonality, The Fourth Subspace, and General Picture of Subspaces
Sep 6, 2020 · 12 min read
1. Recall
(1) Different Subspaces
Let’s say we have an m × n matrix A with A: ℝⁿ → ℝᵐ as
so that we have a null space N(A) is a subspace of ℝⁿ,
and we also have a column space Col(A) is a subspace of ℝᵐ,
and finally, we also have the span of the rows of A as a subspace of ℝⁿ,
It is also important to keep in mind this kind of unfinished graph that we have made in the last section.
(2) Definition: Rank
When we change the matrix A to row echelon form, then we can calculate the rank of A, which is the number of pivots in REF(A), and it is notated by r.
(3) Definition: Basis
A basis for a subspace is a set of linearly independent vectors that span the subspace.
(4) Definition: Dimension
The number of vectors in a basis is called the dimension.
• the basis for Col(A) is given by the pivot columns of the original matrix A
• the basis for the row(A) is given by the nonzero rows of REF(A)
• the basis for N(A) is given by the special solutions to Ax = 0.
• rank nullity theorem
(5) A Quick Example
Suppose we have a 4 x 5 matrix A as follows,
Find the basis for the three subspaces.
Ans:
By row reduction,
then, for the REF(A), we can have,
• rank = 3 = dim Col(A) = dim row(A)
• dim N(A) = 5–3 = 2
Now let’s start with N(A), the free variables are x3 and x4, and to compute N(A), we have to solve Ax = 0. Say x3 = c, x4 = d, then we have the back substitution as,
Finally, we can have the solution x as a result,
or we can also solve this equation by assigning c = 1, d = 0 and c = 0, d = 1.
So finally the vector [-3 -1 1 0 0] and [-7 -3 0 1 0] span the null space N(A), and it is also that the set of {[-3 -1 1 0 0], [-7 -3 0 1 0]} is a basis if this set is linearly independent.
2. The Proofs of Linearly Independent in Subspaces
(1) Recall: The Definition of Linear Independence
A set of vectors {v1, v2, …, vk} in a vector space V is linearly independent provided that,
(a) Perspective 1
whenever,
we must have,
(b) Perspective 2
Another way to think about linear independent is that if I take {v1, v2, …, vk} and put them into a matrix A (as columns) as,
( the size of A is m × k ). Now we have the set {v1, v2, …, vk} is independent ⇔ Ax = 0 has only the trivial solution (x = 0) ⇔ N(A) = {0}
(c) Perspective 3
Another perspective is that because A is m × k, so A: ℝᵏ → ℝᵐ and A is a one-to-one function from ℝᵏ to ℝᵐ. A one-to-one function means that if Av = Aw, then v = w.
(2) Proof of Linear Independence in Null Space
Why those vectors are linearly independent, this is because that when you look back on the coordinates of these vectors, you can find that the only way to make x a trivial solution to 0 is to make c = d = 0, so that there is no way we can get this linear combination a zero without we get the coefficients themselves to be zero.
This is also because the third and fourth coordinates of those variables correspond to the free variables. So those are the c and d choices that we made. So by nature of these choices, if we look at these coordinates right on any linear combinations, the third coordination must be c and the fourth must be d. So that based on the definition of linear independence (perspective 1), we can draw the conclusion that the vectors in the null space must be linearly independent.
(3) Proof of Linear Independence in Column Space
Note that the nullspace tells us about linear independence relations on the columns of A. Because there is a non-trivial solution of Ax = 0. So by the definition of linear independence (perspective 2), we can then draw a conclusion that the column vectors in the matrix A must be linearly independent.
Also note that if someone asks you if a set of vectors is {v1, v2, …, vk} linearly independent, which is also to ask, whether the matrix A consisted by those vectors has a null space N(A) ={0},
You need to make sure that there are k pivots so that N(A) = {0} or dim N(A) = 0, maybe by doing row reduction, and get the row echelon form of A.
Now let’s write down a basis for Col(A). When we talk about Col(A), we have to take pivot columns in A, so we have,
Recall the RREF of A:
It is clear to see that the pivot columns are 1, 2, 5 and they are clearly linearly independent because if any linear combination of these vectors equal to zero, they will have to be the trivial solution and the only none trivial entries in these columns are pivots.
(4) Proof of Why The Span of Pivot Columns is a Basis of the Column Space
We have proved in (7) that all the pivot columns in a matrix are linearly independent. So now if we can also prove that this set is a basis, we have to prove that it spans Col(A).
Because the non-pivot columns (free variables) can be written as linear combinations of the pivot columns, that means we don’t need any non-pivot columns to get a span of A because they are already redundant and will not add anything new.
(5) From Pivot Columns to Non-Povit Columns
There are null space solutions that are precisely telling us what linear combination of pivot columns you need to take to get those non-pivot columns. So it tells us that we don’t actually need those free variable columns.
For example, in the last example, we have one of the non-trivial solutions of Ax = 0 as,
then we multiply the matrix A with this solution as
then we can have,
This means that the column vector a3 can be represented by a linear combination of vector a1 and a2. Thus, we are quite sure that the non-trivial solutions of Ax = 0 actually give us the linear combination of pivot columns you need to take to get those non-pivot columns.
(6) Proof of Linear Independence in Row Space
Finally, we can write down the basis of row(A). The basis of A is given by these non-zero rows of RREF(A). So the basis of the row space should be,
The proof of linear independence of the row space is quite similar to the proof of linear independence of a null space. By the reduced row echelon form, we can find out that the linear combinations of non-zero rows in A should not be 0 unless all the coefficients of this linear combination all equal zero, which is basically the definition of linearly independent in perspective 1.
3. The Orthogonality Between Subspaces
(1) Recall: The Definition of Orthogonality Between Two Vectors
Recall that the vector v and vector w are orthogonal if v · w = 0. This is notated as,
(2) Recall: The Definition of Orthogonality Between Vector And Subspace
Suppose ∀ vector v ∈subspace A, if a given vector w satisfies v · w = 0, then the vector w is orthogonal to subspace A. This is notated as,
(3) Recall: The Definition of Orthogonality Between Two Subspaces
Suppose ∀ vector v ∈subspace A, and ∀ vector wB that satisfies v · w = 0, then the subspace B is orthogonal to subspace A.
In general, if W is a subspace of a vector space V, then the orthogonal subspace of W is defined as,
(4) Orthogonality Between The Null Space and The Column Space of Transpose
In the previous section, we have already proved that the nullspace is the space that perpendicular to the column space of A transpose in the vector space, which is also,
(5) Orthogonality Between The Null Space and The Row Space
Suppose we have vector v and w, and they are orthogonal if v · w = 0. For Ax = 0, we can then observe that,
We have already known that x is the non-trivial solution of Ax = 0, and x is also the basis of null space N(A). It is also clear that by definition, the vector of row in A is also a basis of the row space row(A). Thus, we can say that ∀ x ∈ basis of N(A), we can then have,
Based on the definition of orthogonality, we can observe that,
So for any vector v in the row of A, and any vector x in N(A), we have the property that v · w = 0. Moreover, they have complementary dimensions in ℝⁿ.
(6) Another Perspective of Orthogonality
In the discussion above, we have discovered that both the row space of A ( row(A) ) and the column space of A transpose ( Col(A^T) ) are orthogonal to the null space of A ( N(A) ). This can be explained because the rows in A is equal to the columns in A transpose.
(7) Summary
Based on the discussions above, we can have conclusions that,
and,
4. The Fourth Subspace
We have talked that we want to find a subspace with all its basis vectors ∈ ℝᵐ, and its dimension equals m-r. Could it be possible that we can find a subspace that suffices all those conditions now? Let’s think about A transpose.
(1) A Way to Find the Fourth Space
As we have proved that the row space of A is orthogonal to the null space of A, and they are both subspaces of ℝⁿ, so we can assume that the fourth space is also orthogonal to the column space of A and they are both subspaces of ℝᵐ. Based on the conclusion above, if we use,
then,
So it is quite possible that the nullspace of A transpose is the fourth subspace that we are going to find. Now, it’s time to prove this. We are almost there!
(2) The Definition of The Left-Null Space
By definition of the nullspace, we can know that its basis is the set of all the non-trivial solutions. Suppose the given matrix is A transpose, then the equation that we have to work on is,
and in this case, y is our non-trivial solution.
By the property of transpose,
so,
Based on the form of this expression, the null space of A transpose is sometimes called the left-null space of A.
(3) Proof of Linear Independence
Similarly, as we have said, the linear combination of the basis of non-trivial solutions can be zero only if all the coefficients are zero. Therefore, it is quite clear that all the non-trivial solutions are linearly independent and consists of the basis of the space of A transpose.
(4) The Dimension of The Left-Null Space
By rank nullity theorem, we can have,
then, because the rank of A equals the dimension of the column space of A, then,
it is then clear that,
Thus the left-null space is the fourth subspace that we want to find. So finally, we can add it to our unfinished graph last time.
the fourth subspace that fills in the picture is the nullspace of A transpose, with m-r as the dimension. We can also add the orthogonality onto this graph as,
(5) The basis For Nullspce of A Transpose
Obviously, we have two methods to find a basis here.
• Option 1: do it
The first one is what we have done before. We can first find the transpose of A and then calculate its nullspace by solving the non-trivial solutions. But here we would like to introduce another method to find it.
• Option 2: use row operations to find E · A = REF(A)
For example, we have our A as,
In our case, the A tranpose is 5 × 4 and rank = 3, so dim N(A^T) = 4 –3 = 1.
By doing row operations, we can then have,
if we multiply those elimination matrics together, we can then get an inverse of the lower triangular matrix of A as,
based on the result of the last line, we can then have,
so that it is equivalent to a non-trivial solution of Ax = 0.
(6) General Picture of Subspaces
What is the general solution to Ax = b? Suppose we have xp as a special solution and xn as the linear combination of the non-trivial solutions. We can have a graph as follows,
Then in general, we said the general solution to Ax = b is,
with,
and,
Particularly, if A is n × n with n pivots, then we can always solve Ax = b with a unique solution.
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## The Genetic Algorithm in Solving the Quadratic Assignment Problem
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# If 2x- 5, x-1, and 3x -8 are all integers and x- 1 is the median of these integers, what is x?
Dec 6, 2016
$x = 4$
#### Explanation:
Note first that for all three values to be integers, we must have $x$ be an integer as well.
Because $x - 1$ is the median of the given values, we have
$2 x - 5 \le x - 1 \le 3 x - 8$
or
$3 x - 8 \le x - 1 \le 2 x - 5$.
We consider each case.
Case 1: $2 x - 5 \le x - 1 \le 3 x - 8$
$2 x - 5 \le x - 1$
$\implies 2 x - 5 - x + 5 \le x - 1 - x + 5$
$\implies x \le 4$
$x - 1 \le 3 x - 8$
$\implies x - 1 - x + 8 \le 3 x - 8 - x + 8$
$\implies 7 \le 2 x$
$\implies \frac{7}{2} \le x$
Taken together, we have $x \in \left[\frac{7}{2} , 4\right]$. As $4$ is the only integer in that range, the only solution in this case is $x = 3$.
Case 2: $3 x - 8 \le x - 1 \le 2 x - 5$
If we go through the same steps as above with the directions of the inequalities reverse, we get
$x \ge 4$ and $x \le \frac{7}{2}$. As $\frac{7}{2} < 4$, there are no such values, meaning this case produces no solutions.
Having considered both cases, then, we have found the sole solution as $x = 4$. |
# How to Find a Number of Terms in an Arithmetic Sequence
Co-authored by wikiHow Staff
Updated: September 2, 2019
Finding the number of terms in an arithmetic sequence might sound like a complex task, but it’s actually pretty straightforward. All you need to do is plug the given values into the formula tn = a + (n - 1) d and solve for n, which is the number of terms. Note that tn is the last number in the sequence, a is the first term in the sequence, and d is the common difference.
## Steps
1. 1
Identify the first, second, and last terms of the sequence. Typically, to solve a problem like this, you’ll be given the first 3 or more terms as well as the last term.[1]
• For example, you may have the following sequence: 107, 101, 95…-61. In this case, the first term is 107, the second term is 101, and the last term is -61. You need all of this information to solve the problem.
2. 2
Subtract the first term from the second term to find the common difference. In the example sequence, the first term is 107 and the second term is 101. So, subtract 107 from 101, which is -6. Therefore, the common difference is -6.[2]
3. 3
Use the formula tn = a + (n - 1) d to solve for n. Plug in the last term (tn), the first term (a), and the common difference (d). Work through the equation until you’ve solved for n.[3]
• For example, start by writing: -61 = 107 + (n - 1) -6. Subtract 107 from both sides so you’re left with -168 = (n - 1) -6. Then, divide both sides by -6 to get 28 = n - 1. Finish by adding 1 to both sides so that n = 29.
## Community Q&A
Search
• Question
Can I use the formula (A - L/d) + 1 for finding n?
Donagan
As explained above, n = [(L - A) / d] + 1.
• Question
If the difference is not given, how do I find the difference?
Donagan
Subtract any term from the one that follows it.
• Question
If the first term and last term of an arithmetic progression are 5 and 89, how do I find the number of terms?
Donagan
You can't without knowing the difference between consecutive terms.
• Question
What is the common difference in a minus?
Here is an example: -2, -5, -8, -11. In this sequence the common difference (d) is found by subtracting any term from the term that precedes it. For instance, -5 - (-2) = -5 + 2 = -3. The common difference is a negative 3, which means 3 is always subtracted from a given term to find the next term in the sequence.
• Question
Does an arithmetic sequence start at n=0?
When given a sequence, you'd usually get the first term. Unless n(1)=0, then no, the arithmetic sequence doesn't always begin with 0.
• Question
How do I find the term number? Can someone explain in an easy, simple way?
Donagan
To find the number of terms in an arithmetic sequence, divide the common difference into the difference between the last and first terms, and then add 1.
• Question
How can I find number of items when arithmetic mean is given but sum of items is not given?
Donagan
You would need more information, such as the common difference and the first and last terms.
• Question
How many terms are there in 31, 32, 33, 47, 48, 49?
Assuming this sequence consists of every integer from 31 through 49, subtract 31 from 49, then add 1.
• Question
I've been given this sequence: 2+10+18. How do I find the sum of the first 40 terms?
Donagan
Finding the sum of an arithmetic sequence involves finding the average of the first and last numbers of the sequence. Therefore, you must know the 40th term. Once you find the 40th term (there's a wikiHow article on finding a certain term in an arithmetic sequence), add it to 2, divide by 2, then multiply by 40. That's the sum you're looking for.
• Question
I've been given the sum of the sequence, the first term, and the common difference. How can I get the number of terms?
Donagan
You don't have enough information to find the number of terms quickly. However, you could start at the first term and keep adding the common difference over and over until you reach the given sum of the sequence. Count the number of times you added the common difference. Add 1 to that number to get the number of terms in the sequence.
• How do I get and solve the first term in an arithmetic sequence?
• What does a '1' represent in the formula?
• If the sum of an arithmetic progression is 200, and if the first term is 20, what is the total number of terms in this arithmetic progression?
• What is the value of n for which the nth term is 675?
• I've been given the sum of the sequence, the last term and the common difference. How can I determine the total number of terms?
200 characters left
## Tips
• The difference between the last term and first term will always be divisible by the common difference.
Thanks!
Submit a Tip
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## Warnings
• Do not confuse the difference between the first and last term with the common difference.
Thanks!
Co-Authored By:
wikiHow Staff Editor
This article was co-authored by our trained team of editors and researchers who validated it for accuracy and comprehensiveness.
Co-authors: 9
Updated: September 2, 2019
Views: 201,060
Categories: Algebra
Thanks to all authors for creating a page that has been read 201,060 times.
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Reduce / simplify any portion to that lowest state by utilizing our portion to the Simplest kind Calculator. Discover the prize to questions like: What is 15/45 in simplest form or what is 15/45 diminished to the simplest form?
### Fractions Simplifier
Please to fill in the two left crate below:
Inputfraction: Integerpart: Fractionpart: As a Decimal: = = Details:Details...You are watching: What is 15/45 simplified
## How to reduce a fraction
Among different ways simple a fraction, us will show the 2 procedure below:
### Method 1 - divide by a small Number as soon as Possible
Start by separating both the numerator and also the denomiator of the fraction by the same number, and repeat this until it is impossible to divide. Start dividing by little numbers choose 2, 3, 5, 7. Because that example,
### Simplify the portion 42/98
First divide both (numerator/denominator) by 2 to acquire 21/49.Dividing through 3 and 5 will not work, so,Divide both numerator and denominator through 7 to gain 3/7. Note: 21 ÷ 7 = 3 and 49 ÷ 7 = 7
In the fraction 3/7, 3 is only divisible through itself, and also 7 is not divisible by other numbers than itself and 1, therefore the fraction has been simplified as much as possible. No further reduction is possible, for this reason 42/98 is equal to 3/7 when reduced to its lowest terms. This is a PROPER portion once the absolute worth of the top number or numerator (3) is smaller sized than the absolute worth of the bottom number or denomintor (7).
### Method 2 - Greatest usual Divisor
To reduce a portion to lowest state (also dubbed its most basic form), just divide both the numerator and also denominator by the GCD (Greatest usual Divisor).
For example, 3/4 is in lowest form, yet 6/8 is not in lowest type (the GCD the 6 and 8 is 2) and 6/8 have the right to be created as 3/4. You have the right to do this since the value of a portion will stay the same when both the numerator and denominator are separated by the same number.
Note: The Greatest usual Factor (GCF) because that 6 and 8, notation gcf(6,8), is 2. Explanation:
Factors the 6 space 1,2,3,6;Factors that 8 are 1,2,4,8.
See more: The Basic Landform Pattern Of Central America Consists Of, Geo Chapter 11 Questions Flashcards
So, it is ease view that the "Greatest common Factor" or "Divisor" is 2 due to the fact that it is the biggest number which divides same into every one of them. |
# The elementary mathematics of binary-decimal converters
If you google "How to convert from decimal to binary," you'll find four easy algorithms, two for the integer and two for fractions. In the first section of this article, we provide some examples of each. Even though it's usually sufficient to merely be familiar with the algorithms, I've made it a goal to learn how and why they function. Section 2 of this article delves into the elementary mathematics behind each of these. If you ever find yourself forgetting an algorithm, knowing this might help you recall it. You should get a notepad and a pen and follow along with the calculations as I do them to help you remember the material. You can find examples of each of the four algorithms below on the internet.
The first step in converting an integer to binary is to divide it by 2 while keeping track of the quotient and the remainder. Repeat dividing by 2 until the quotient becomes zero. The remainders can then be written in descending order.
Take the number 12 as an illustration of such a transformation.
We'll start by dividing the number in half and writing down the quotient and remainder.
The rest can be written in reverse order (1100) now. Therefore, in binary, the number 12 corresponds to 1100 in the decimal system.
Multiply the fraction by 2, keeping track of the integer and fractional part, and you'll have a binary representation for the fraction. Keep multiplying by two until the resulting fraction is zero. Then, simply record the integer factors resulting from each multiplication.
See the fraction 0 as an example of a number that can be converted to decimal form below. 375
Let's just output the resulting integer fraction at each iteration, which is 0 in this case. 011 So, 0 There is no digit for 375 in the decimal system. This is the binary representation of the number 11.
The only fractions that can be represented exactly in binary form have denominators that are powers of two. Examples of this include denominators of zero and similar. 1 (1 / 10) and 0 Binary representations are limited to powers of two, but 2 and 15 are not powers of two. Mantissa bits must be rounded down to fit into the IEEE-754 floating point representation, which is 10 bits for half-precision, 23 bits for single-precision, and 52 bits for double-precision. Floating-point approximations of 0 vary in accuracy from one implementation to the next. 1 and 0 Numbers written in decimal form can be close to 2, but never exactly 2. That's why you'll never be at a loss for words and never reach zero. 1 0 2 == 0 3
Begin with the left when performing a left-to-right conversion of binary integers to decimal. Multiply the number you're adding to by two, and add the number you're adding. Keep going until no more digits remain. The fraction 1011 is used here to illustrate a similar conversion.
When converting a binary fraction to a decimal form, zero is the starting point. To get the next digit, add the current one, and divide by 2; apply this to your current total. Keep going until there are no more digits An illustration of such a transformation using the fraction 0 is provided below. 1011 To do this, I've simply substituted multiplication by 1/2 for division by 2.
You can now convert binary numbers to and from decimal with the help of these four straightforward algorithms.
The key to comprehending the efficacy of those algorithms lies in the number's expansion in base q. An integer in any numeral system looks like this:
where,
• The value of N is a whole number.
• The x represents the digit (0–9 in the base–10 system; 0–1 in the base–2 system).
• It all starts with q, where 10 in our case is the base value for our system and 2 is the base value for our base-2 system.
Base q expansion of the number N, or just base q expansion, will be used throughout this article. First, let's compare the binary and decimal representations of the number 12.
Similarly, any fractional number can be written as follows, regardless of the system used.
where,
• N is a decimal number
• The x represents the digit (0–9 in the base–10 system; 0–1 in the base–2 system).
• In a base-10 system, q is 10, and in a base-2 system, q is 2.
Referring to the digit zero 375 is written as follows in both the binary and decimal systems.
A surprising application of this base-q expansion form is the conversion of a decimal number to binary. We can do the same thing for the number 12: Before delving into the binary representation, let's pretend we have no idea and write it out with xs in place of the unknown digits.
It is our job to track down every x. So, let's see what we can do. All summands, with the exception of the last, are multiples of two, so they are all even numbers. Based on this, we can determine the value of x0: if the input integer is even, then x0 is 0, and if it is odd, then x0 is 1. Since 12 is an even number, x0 here is also zero. Yes, let's put this down on paper:
The x1 value must be determined next. Factoring out 2 allows us to isolate x1, as all summands from x1 to xN are multiples of 2. In that case, let's:
It is also evident that the total of the values contained within the brackets is 6. As a result, the first actionable item is:
To continue determining the missing x's, please. We can express the polynomial within the brackets as a standalone statement:
The same reasoning used up above allows us to confirm that x1 = 0. Let’s rewrite it and take 2 into account once more:
Then, the second step is:
There is a pattern emerging now. After the initial factorization, we can keep factoring 2 until the quotient becomes 0. Okay, so let's keep this pattern going and see what happens.
Given that the quotient is 1, we now have only one remaining summand and can rewrite the expression as:
That brings us to our third and final step:
The resulting scenario is as follows:
There is no doubt that x3 = 1. However, since a quotient is required by our algorithm, let's rewrite the previous expression as follows:
When we divide by zero, we're left with nothing and have completed the process. How about we put it in writing?
Therefore, the process of conversion is complete. We've laid out the stages of our transformation below:
It is now obvious that the value of the remainder at each step is equivalent to the value of the x at the corresponding position. According to the aforementioned algorithm, the binary representation of the number 12 is 1100.
It's important to recall that our original goal was to demonstrate the efficacy of the algorithm involving a "diving by 2" Follow the above procedure and shift the 2 to the left side of the expressions:
Here you can see the logical progression that led to the algorithm described in the introduction. The numbers for all four procedures can be represented in this way, as well.
If you want to learn about the algorithm for converting binary to decimal, you'll need to know how we arrive at that representation.
I'll also use the base-q expansion form of fractions to demonstrate why we multiply by 2 and take the integer part when translating between the two systems. As a decimal fraction, I'll use 0. the first 375 words of the article In the same vein as the integer portion, let's pretend we don't know the binary representation of this number and write it out with xs standing in for the unknown digits:
The goal, as with integers, is to identify every instance of x. So, let's figure out a way to make that happen. The first thing to note is that when we divide by a negative power of 2, we get a fraction where the denominator is 2. Change the above phrase into the following:
The solution is as obvious as factoring out 1/2 in the appropriate part of the expression. Sure, let's do that:
Half a step to the left, and we're good to go!
We now have a single variable, x1, which can take on the values 1 or 0. What digit it has can be found by examining the remaining summands:
Consider how large this set of numbers can potentially become. Simply replace all x's with 1's to get the final sum if the greatest possible value for x is 1.
Since the sum of a geometric series of fractions is constrained to fall within the interval [0, sum], the greatest number that such a sum can yield is 1. Let's take another look at our idiom:
It should be obvious that if the right side is less than 1, then x1 can't be equal to 1, and so it's equal to 0. The rest of the expression is also equal to 0. 75
This matches the initial description of the algorithm perfectly:
First, let's decimalize 0 to make it more manageable. 75.5, then subtract 1/2 to get x2 all by itself:
as well as shift 1/2 to the left:
The left side of the expression cannot be greater than 1 if x2 is zero, but it is one. 5.x2 = 1, leaving the rest to be 0. 5 So, to put it in writing:
And this, once again, conforms to the scheme introduced at the outset:
For the remainder of the value of zero, let's proceed in the same manner. 5
The same reasoning as before reveals that x3 = 1 and leaves no remainder in the fraction:
Final step looks like this since the remaining fractional part is 0.
So, let's re-write the entire procedure:
My original algorithm is exactly like this. The computations for all three steps can be represented in a single form, just as we did with integers:
Once again, a thorough understanding of this representation is crucial, as we will be using it to investigate binary to decimal conversion.
Many programmers are surprised to learn that some fractions that can be represented finitely in the decimal system cannot be represented finitely in binary. It is this misunderstanding, however, that underlies the seemingly strange result obtained by adding zeros. 1 to 0 2 What then is it about a fraction that decides whether or not it can be represented by a finite number system? It's a lot of work If you want to keep things simple, just remember that the denominator of a fraction should be a power of the system base for any number to be represented finitely. In order to represent zero exactly in the base-10 system, the denominator must be a power of 10. In decimal form, the number 625 reads as follows:
and cannot be represented exactly as 1/3:
The same applies to base-2 numbers:
However, if we rule out the number 0, 1 because 10 is not a power of 2 and therefore 0 is the answer. Within the framework of the binary system, 1 will represent an infinite fraction. Applying the above-mentioned algorithm, let's take a look:
This process can be repeated indefinitely, but for the sake of clarity, let's express it as a continued fraction.
In order to demonstrate why the algorithm of multiplying by 2 is effective, I will once again use the binary integer 1011. The number will be written in its base-q expansion form here as well. Let's put it in this form for easy reference:
Due to the fact that all summands are divisible by 2, we can factor out 2 until the quotient becomes 0. Come on, let's do it:
Now, if you just use the order of operations in math, you'll get the same results I showed at the outset, including:
The binary representation of the number 11 is 1011.
Finally, we've reached the end of the algorithms! You've probably figured out the workings of it on your own by now. If that's not the case, then let's examine the logic behind its success. The key is again the expanded form of the number in base q. To start, let's pretend that 0 is our starting point. 1011, from the introduction Let’s put it in its full form:
Since every summand is a power of 2, we can continue to remove 1/2 as a factor until there is no longer any fractional component. Yes, let's do that.
The initialized algorithm is the result of performing the mathematical operations in the specified order:
This way, 0 The binary representation of the number 1011 is 0. That's the decimal equivalent of "6875"
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# Video: Finding an Unknown Element Using the Relation between Two Given Sets
Given that {8, 7} ⊂ {9, 8, 𝑥}, find the value of 𝑥.
03:10
### Video Transcript
Given that a set containing the elements eight and seven is a subset of a set containing the elements nine, eight, and 𝑥, find the value of 𝑥.
In this problem, we’ve got two sets that are being described. Let’s try and represent them using circles. The first set contains the elements eight and seven. And we can see them written there in between the braces with a comma in between them. Our second set contains three elements: nine, eight, and the letter 𝑥, which we know represents a number. We know this because often in algebra, the letter 𝑥 represents numbers. But also in this particular question, we’re asked to find the value of 𝑥.
So, here are our two sets. What relationship do they have to each other? Well, in between the notation for both sets, we have this symbol. Let’s remind ourselves what it means. Well, this symbol, which looks a bit like a stretched-out letter C, means is a subset of. Now, when a set is a subset of another set, it’s contained within it. So, if we said A is a subset of B, it’s contained within B. It’s part of B.
Let’s have another look at the sets in our question. We’re told that a set containing the elements eight and seven is a subset of, or is part of, a set containing the elements nine, eight, and 𝑥. Let’s try drawing our circles again to reflect this. This time, we can draw our second set first. It contains the elements, remember, nine, eight, and 𝑥.
But we’re told that a set containing the elements eight and seven is a subset of this. Well, we should have no problem understanding where the number eight comes from. We can see that in the second set. But where is the number seven? How could we make a subset containing eight and seven when all we have are the digits nine, eight, and of course the letter 𝑥?
Well, the answer is, of course, that the letter 𝑥 must have a value. And we can see what this value is because we know what the subset of this set is. Let’s draw a circle within this set to show the subset. So, now we can write that a set containing the elements eight and 𝑥 is a subset of a set containing the elements nine, eight, and 𝑥. And so, if 𝑥 equals seven, then the statement in the question is still true. If a set containing the elements eight and seven is a subset of a set containing the elements nine, eight, and 𝑥, then we can say that 𝑥 equals seven. |
Class 8 Maths Factorisation What is Factorization
What is Factorization
Factors of a number are numbers that divide evenly into another number. Factorization writes a number
as the product of smaller numbers.
For example, let’s factor the number 12:
12 = 6 * 2 = 3 * 4 = 2 * 2 * 3
Again 3xy + 2x, x2 + 10x, x2 + 7x + 12, etc are not in the factor form.
Method of Common Factors:
Let us take an example.
Ex: Factorize 3x + 9
Now, we write each term as a product of irreducible factors.
3x = 3 * x
9 = 3 * 3
So, 3x + 9 = (3 * x) + (3 * 3)
= 3 * (x + 3) [By Distributive Law]
= 3(x + 3)
Again let we factorize 10xy + 2y
Now, the irreducible factor forms of 10xy and 4y are respectively,
10xy = 2 * 5 * x * y
2y = 2 * 2 * y
10xy + 2y = (2 * 5 * x * y) + (2 * 2 * y)
= (2y * 5x) + (2y * 2)
= 2y * (5x + 2) [By Distributive Law]
= 2y(5x + 2)
Factorization by regrouping terms:
Let we want to factorize x2 + xy + 8x + 8y
Since there is no common factor among all terms. So, we form groups.
We form two groups (x2 + xy) and (8x + 8y) and then factorize
x2 + xy = x * x + x * y
= x * (x + y)
8x + 8y = 8 * x + 8 * y
= 8 * (x + y)
Now, x2 + xy + 8x + 8y = x * (x + y) + 8 * (x + y)
= (x + y)(x + 8)
We can regroup the algebraic expressions in many ways.
Problem: Find the common factors of the given terms.
(i) 12x, 36 (ii) 2y, 22xy (iii) 14pq, 28 p2 q2 (iv) 2x, 3x2, 4
(i) 12x = 2 * 2 * 3 * x
36 = 2 * 2 * 3 * 3
Hence, the common factors are 2, 2 and 3 = 2 * 2 * 3 = 12
(ii) 2y = 2 * y
22xy = 2 * 11 * x * y
Hence, the common factors are 2 and y = 2 * y = 2y
(iii) 14pq * 2 * 7 * p * q
28 p2 q2 = 2 * 2 * 7 * p * p * q * q
Hence, the common factors are 2 * 7 * p * q = 14pq
(iv) 2x =2 * x * 1
3x2 = 3 * x * x * 1
4 = 2 * 2 * 1
Hence, the common factors is 1
Problem: Factorize:
(i) 15xy – 6x + 5y – 2 (ii) ax + bx – ay – by (iii) 15pq + 15 + 9q + 25p (iv) z – 7 + 7xy – xyz
(i) 15xy – 6x + 5y – 2 = 3x(5y - 2) + 1(5y - 2)
= (5y - 2)(3x + 1)
(ii) ax + bx – ay – by = x(a + b) – y(a + b)
= (a + b)(x - y)
(iii) 15pq + 15 + 9q + 25p = 15pq + 25p + 15 + 9q
= 5p(3q + 5) + 3(3q + 5)
= (3q + 5)(5p + 3)
(iv) z – 7 + 7xy – xyz = 7xy – 7 – xyz + z
= 7(xy - 1) – z(xy - 1)
= (xy - 1)(7 - z)
Factorization using Identity:
We know the identity
(a + b)2 = a2 + 2ab + b2
(a - b)2 = a2 - 2ab + b2
(a + b)(a - b) = a2 – b2
(x + a)(x + b) = x2 + (a + b)x + ab
We use these identities to do the factorization of algebraic expressions.
Problem: Factorize the following expressions:
(i) a2 + 8a + 16 (ii) p2 – 10p + 25 (iii) 25m2 + 30m + 9
(iv) 49y2 + 84yz + 36z2 (v) 4x2 – 8x + 4 (vi) 121b2 – 88bc + 16c2
Solution:
(i) a2 + 8a + 16 = a2 + (4 + 4)a + 4 * 4
= (a + 4)(a + 4) [x2 + (a + b) + ab = (x + a)(x + b)]
= (a + 4)2
(ii) p2 – 10p + 25 = p2 – (-5 - 5)p + (-5)(-5)
= (p - 5)(p - 5)
= (p - 5)2 [x2 + (a + b) + ab = (x + a)(x + b)]
(iii) 25m2 + 30m + 9 = (5m)2 + 2 * 5m * 3 + 32
= (5m + 3)2 [a2 + 2ab + b2 = (a + b)2]
(iv) 49y2 + 84yz + 36z2 = (7y)2 + 2 * 7y * 6z + (6z)2
= (7y + 6z)2 [a2 + 2ab + b2 = (a + b)2]
(v) 4x2 – 8x + 4 = (2x)2 – 2 * 2x * 2 + 22
= (2x - 2)2 [a2 - 2ab + b2 = (a - b)2]
= 22(x - 1)2
= 4(x - 1)2
(vi) 121b2 – 88bc + 16c2 = (11b)2 – 2 * 11b * 4c + (4c)2
= (11b – 4c)2 [a2 - 2ab + b2 = (a - b)2]
. |
# How to Solve Natural Logarithms Problems? (+FREE Worksheet!)
In this blog post, you will learn more about Natural Logarithms and how to solve problems related to natural logarithms.
## Step by step guide to solve Natural Logarithms
• A natural logarithm is a logarithm that has a special base of the mathematical constant $$e$$, which is an irrational number approximately equal to $$2.71$$.
• The natural logarithm of $$x$$ is generally written as ln $$x$$, or $$\log_{e}{x}$$.
### Natural Logarithms – Example 1:
Solve the equation for $$x$$: $$e^x=3$$
Solution:
If $$f(x)=g(x)$$,then: $$ln(f(x))=ln(g(x))→ln(e^x)=ln(3)$$
Use log rule: $$\log_{a}{x^b}=b \log_{a}{x}$$, then: $$ln(e^x)=x ln(e)→xln(e)=ln(3)$$
$$ln(e)=1$$, then: $$x=ln(3)$$
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### Natural Logarithms – Example 2:
Solve equation for $$x$$: $$ln(2x-1)=1$$
Solution:
Use log rule: $$a=\log_{b}{b^a}$$, then: $$1=ln(e^1 )=ln(e)→ln(2x-1)=ln(e)$$
When the logs have the same base: $$\log_{b}{f(x)}=\log_{b}{g(x)}$$, then: $$f(x)=g(x)$$
then: $$ln(2x-1)=ln(e)$$, then: $$2x-1=e→x=\frac{e+1}{2}$$
### Natural Logarithms – Example 3:
Solve the equation for $$x$$: $$e^x=5$$
Solution:
If $$f(x)=g(x)$$,then: $$ln(f(x))=ln(g(x))→ln(e^x)=ln(5)$$
Use log rule: $$\log_{a}{x^b}=b \log_{a}{x}$$, then: $$ln(e^x)=x ln(e)→xln(e)=ln(5)$$
$$ln(e)=1$$, then: $$x=ln(5)$$
### Natural Logarithms – Example 4:
Solve equation for $$x$$: $$ln(5x-1)=1$$
Solution:
Use log rule: $$a=\log_{b}{b^a}$$, then: $$1=ln(e^1 )=ln(e)→ln(5x-1)=ln(e)$$
When the logs have the same base: $$\log_{b}{f(x)}=\log_{b}{g(x)}$$, then: $$f(x)=g(x)$$
then: $$ln(5x-1)=ln(e)$$, then: $$5x-1=e→x=\frac{e+1}{5}$$
## Exercises to practice Natural Logarithms
The Perfect Book to Ace the College Algebra Course
### Solve each equation for $$x$$.
1. $$\color{blue}{e^x=3}$$
2. $$\color{blue}{e^x=4}$$
3. $$\color{blue}{e^x=8}$$
4. $$\color{blue}{ln x=6}$$
5. $$\color{blue}{ln (ln x)=5}$$
6. $$\color{blue}{e^x=9}$$
7. $$\color{blue}{ln(2x+5)=4}$$
8. $$\color{blue}{ln(2x-1)=1}$$
• $$\color{blue}{x=ln 3}$$
• $$\color{blue}{x=ln 4,x=2ln(2)}$$
• $$\color{blue}{x=ln 8,x=3ln(2)}$$
• $$\color{blue}{x=e^6}$$
• $$\color{blue}{x=e^{e^5}}$$
• $$\color{blue}{x=ln 9,x=2ln(3)}$$
• $$\color{blue}{x=\frac{e^4-5}{2}}$$
• $$\color{blue}{x=\frac{e+1}{2}}$$
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# What is the equation of the line that passes through (-1,2) and is perpendicular to the line that passes through the following points: (5,-3),(7,6) ?
Apr 15, 2016
$y = \frac{- 2 x}{9} + \frac{16}{9}$
#### Explanation:
The general method to solving questions of this nature is this:
1. Find the slope of the line between the last two points
2. From that find slope of final line in answer
3. Deduce equation from the calculated slope and other point
Step 1:
We can find the slope between lines $\left(5 , - 3\right) \mathmr{and} \left(7 , 6\right)$ using the gradient formula:
$m = \frac{y 2 - y 1}{x 2 - x 1}$
$m = \frac{6 - \left(- 3\right)}{7 - 5}$
$m = \frac{9}{2}$
Step 2:
We can find the gradient of the line we're looking for by using this equation:
$m 1 = - \frac{1}{m 2}$ , where m2 would be $\frac{9}{2}$ and m1 would be the gradient we're looking for. Therefore:
$m 1 = - \frac{1}{\frac{9}{2}}$
$m 1 = - \frac{2}{9}$
Step 3:
Knowing the gradient of the line and a point that passes through it, we can use the point gradient formula to get the final answer:
$y - y 1 = m \left(x - x 1\right)$
$y - 2 = - \frac{2}{9} \left(x + 1\right)$
$y - 2 = \frac{- 2 x}{9} - \frac{2}{9}$
$y = \frac{- 2 x}{9} + \frac{16}{9}$ |
# How do you simplify square root of 3x + 3 square root of 3x - 2 square root of 3x - 10?
Oct 8, 2015
$2 \cdot \left(\sqrt{3 x} - 5\right)$
#### Explanation:
Assuming that your expression looks like this
$\sqrt{3 x} + 3 \sqrt{3 x} - 2 \sqrt{3 x} - 10$
you can use $\sqrt{3 x}$ as a common factor to get
$\sqrt{3 x} \cdot \left(1 + 3 - 2\right) - 10$
This will be equal to
$2 \sqrt{3 x} - 10$
Finally, use $2$ as a common factor to get the simplified form of the expression
$2 \sqrt{3 x} - 10 = \textcolor{g r e e n}{2 \cdot \left(\sqrt{3 x} - 5\right)}$ |
# Student t Distribution: Definition & Example
Coming up next: Using the t Distribution to Find Confidence Intervals
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• 0:05 T Distribution
• 0:44 Degrees of Freedom
• 1:37 Important Properties
• 2:15 Example
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Lesson Transcript
Instructor: Artem Cheprasov
In this lesson, you're going to learn about the t-distribution, t-curves, their important properties, and differences from the standard normal distribution as well as how to find the value of t.
## T Distribution
The t distribution also known as the Student's t distribution is a kind of symmetric, bell-shaped distribution that has a lower height but a wider spread than the standard normal distribution. The units of a t distribution are denoted with a lower case 't'.
If you look at the image on your screen, you can see how each corresponding curve is bell-shaped and are symmetric about 0, but the t distribution has a bigger spread than the standard normal distribution curve, in essence the t distribution has a large standard deviation.
A standard deviation refers to the variability of individual observations around their mean.
Let's learn a bit more about the t distribution.
## Degrees of Freedom
The only parameter of the t distribution is the number of degrees of freedom. The degrees of freedom (df) are simply n-1. Meaning df = n - 1, where n is our sample size.
The shape of each individual t distribution curve depends on the degrees of freedom, but all t-curves still resemble the standard normal curve nonetheless. Why does a t-curve have more spread than the standard normal curve?
It's because the standard deviation for a t-curve with v degrees of freedom, where v > 2, is the square root of v divided by v - 2. Because this value is always greater than 1, which is the standard deviation of the standard normal distribution curve, the spread is thus larger for a t-curve.
## Important Properties
There are several important properties you should be aware of with respect to t-curves.
Property #1: The total area under a t distribution curve is 1.0: that is 100%.
Property #2: A t-curve is symmetric around 0.
Property #3: While a t-curve extends infinitely in either direction, it approaches, but never touches the horizontal axis.
Property #4: As the number of df increases, the t distribution curve will look more and more like the standard normal distribution curve.
## Example
Let's solidify our knowledge of the t distribution with an example. Using 15 degrees of freedom and a 0.05 area in the right tail of a t-curve, find the value of t.
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## How do I involve my child in challenging mathematics?
“How do I involve my child in challenging mathematics? He gets good marks in school tests but I think he is smarter than school curriculum.”
“My daughter is in 4th grade. What competitions in mathematics and science can she participate in? How do I help her to perform well in those competitions?”
“I have a 6 years old kid. He hates math. How do I change that?”
We often get queries and requests like these from parents around the world. Literally. In fact, the first one came from Oregon, United States, the second one from Cochin, India, and the last one from Singapore.
Categories
## Initiating a child into the world of Mathematical Science
“How do I involve my son in challenging mathematics? He gets good marks in school tests but I think he is smarter than school curriculum.”
“My daughter is in 4th grade. What competitions in mathematics and science can she participate? How do I help her to perform well in those competitions?”
“I have a 6 years old kid. He hates math. How do I change that?”
We often get queries and requests like these from parents around the world. Literally. In fact the first one came from Oregon, United States, second one from Cochin, India and last one from Singapore.
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## AMC 10 (2013) Solutions
12. In $(\triangle ABC, AB=AC=28)$ and BC=20. Points D,E, and F are on sides $(\overline{AB}, \overline{BC})$, and $(\overline{AC})$, respectively, such that $(\overline{DE})$ and $(\overline{EF})$ are parallel to $(\overline{AC})$ and $(\overline{AB})$, respectively. What is the perimeter of parallelogram ADEF?
$(\textbf{(A) }48\qquad\textbf{(B) }52\qquad\textbf{(C) }56\qquad\textbf{(D) }60\qquad\textbf{(E) }72\qquad )$
Solution: Perimeter = 2(AD + AF). But AD = EF (since ABCD is a parallelogram).
Hence perimeter = 2(AF + EF).
Now ABC is isosceles (AB = AC = 28). Thus angle B = angle C. But EF is parallel to AB. Thus angle FEC = angle B which in turn is equal to angle C.
Hence triangle CEF is isosceles. Thus EF = CF.
Perimeter = 2(AF + EF) = 2(AF + EF) =2AC = $(2 \times 28)$ = 56.
Ans. (C) 56
Categories
## USAJMO 2012 questions
1. Given a triangle ABC, let P and Q be the points on the segments AB and AC, respectively such that AP = AQ. Let S and R be distinct points on segment BC such that S lies between B and R, ∠BPS = ∠PRS, and ∠CQR = ∠QSR. Prove that P, Q, R and S are concyclic (in other words these four points lie on a circle).
2. Find all integers $(n \ge 3 )$ such that among any n positive real numbers $( a_1 , a_2 , ... , a_n )$ with $\displaystyle {\text(\max)(a_1 , a_2 , ... , a_n) \le n) (\min)(a_1 , a_2 , ... , a_n)}$ there exist three that are the side lengths of an acute triangle.
3. Let a, b, c be positive real numbers. Prove that $\displaystyle {(\frac{a^3 + 3 b^3}{5a + b} + \frac{b^3 + 3c^3}{5b +c} + \frac{c^3 + 3a^3}{5c + a} \ge \frac{2}{3} (a^2 + b^2 + c^2))}$.
4. Let $(\alpha)$ be an irrational number with $(0 < \alpha < 1)$, and draw a circle in the plane whose circumference has length 1. Given any integer $(n \ge 3 )$, define a sequence of points $(P_1 , P_2 , ... , P_n )$ as follows. First select any point $(P_1)$ on the circle, and for $( 2 \le k \le n )$ define $(P_k)$ as the point on the circle for which the length of the arc $(P_{k-1} P_k)$ is $(\alpha)$, when travelling counterclockwise around the circle from $(P_{k-1} )$ to $(P_k)$. Suppose that $(P_a)$ and $(P_b)$ are the nearest adjacent points on either side of $(P_n)$. Prove that $(a+b \le n)$.
5. For distinct positive integers a, b < 2012, define f(a, b) to be the number of integers k with (1le k < 2012) such that the remainder when ak divided by 2012 is greater than that of bk divided by 2012. Let S be the minimum value of f(a, b), where a and b range over all pairs of distinct positive integers less than 2012. Determine S.
6. Let P be a point in the plane of triangle ABC, and $(\gamma)$ be a line passing through P. Let A’, B’, C’ be the points where reflections of the lines PA, PB, PC with respect to $(\gamma)$ intersect lines BC, AC, AB, respectively. Prove that A’, B’ and C’ are collinear. |
# Mathematics | Euler and Hamiltonian Paths
Prerequisite – Graph Theory Basics
Certain graph problems deal with finding a path between two vertices such that each edge is traversed exactly once, or finding a path between two vertices while visiting each vertex exactly once. These paths are better known as Euler path and Hamiltonian path.
The Euler path problem was first proposed in the 1700’s.
Euler paths and circuits :
• An Euler path is a path that uses every edge of a graph exactly once.
• An Euler circuit is a circuit that uses every edge of a graph exactly once.
• An Euler path starts and ends at different vertices.
• An Euler circuit starts and ends at the same vertex.
The Konigsberg bridge problem’s graphical representation :
There are simple criteria for determining whether a multigraph has a Euler path or a Euler circuit. For any multigraph to have a Euler circuit, all the degrees of the vertices must be even.
Theorem – “A connected multigraph (and simple graph) with at least two vertices has a Euler circuit if and only if each of its vertices has an even degree.”
Proof of the above statement is that every time a circuit passes through a vertex, it adds twice to its degree. Since it is a circuit, it starts and ends at the same vertex, which makes it contribute one degree when the circuit starts and one when it ends. In this way, every vertex has an even degree.
Since the Koningsberg graph has vertices having odd degrees, a Euler circuit does not exist in the graph.
Theorem – “A connected multigraph (and simple graph) has an Euler path but not an Euler circuit if and only if it has exactly two vertices of odd degree.”
The proof is an extension of the proof given above. Since a path may start and end at different vertices, the vertices where the path starts and ends are allowed to have odd degrees.
• Example – Which graphs shown below have an Euler path or Euler circuit?
• Solution – has two vertices of odd degree and and the rest of them have even degree. So this graph has an Euler path but not an Euler circuit. The path starts and ends at the vertices of odd degree. The path is- .
has four vertices all of even degree, so it has a Euler circuit. The circuit is – .
Hamiltonian paths and circuits :
Hamilonian Path – A simple path in a graph that passes through every vertex exactly once is called a Hamiltonian path.
Hamilonian Circuit – A simple circuit in a graph that passes through every vertex exactly once is called a Hamiltonian circuit.
Unlike Euler paths and circuits, there is no simple necessary and sufficient criteria to determine if there are any Hamiltonian paths or circuits in a graph. But there are certain criteria which rule out the existence of a Hamiltonian circuit in a graph, such as- if there is a vertex of degree one in a graph then it is impossible for it to have a Hamiltonian circuit.
There are certain theorems which give sufficient but not necessary conditions for the existence of Hamiltonian graphs.
Dirac’s Theorem- “If is a simple graph with vertices with such that the degree of every vertex in is at least , then has a Hamiltonian circuit.”
Ore’s Theorem- “If is a simple graph with vertices with such that for every pair of non-adjacent vertices and in , then has a Hamiltonian circuit.”
As mentioned above that the above theorems are sufficient but not necessary conditions for the existence of a Hamiltonian circuit in a graph, there are certain graphs which have a Hamiltonian circuit but do not follow the conditions in the above-mentioned theorem. For example, the cycle has a Hamiltonian circuit but does not follow the theorems.
Note: Kn is Hamiltonian circuit for
There are many practical problems which can be solved by finding the optimal Hamiltonian circuit. One such problem is the Travelling Salesman Problem which asks for the shortest route through a set of cities.
• Example 1- Does the following graph have a Hamiltonian Circuit?
• Solution- Yes, the above graph has a Hamiltonian circuit. The solution is –
• Example 2- Does the following graph have a Hamiltonian Circuit?
• Solution- No the above graph does not have a Hamiltonian circuit as there are two vertices with degree one in the graph.
GATE CS Corner Questions
Practicing the following questions will help you test your knowledge. All questions have been asked in GATE in previous years or in GATE Mock Tests. It is highly recommended that you practice them.
References-
Eulerian path – Wikipedia
Hamiltonian path – Wikipedia
Discrete Mathematics and its Applications, by Kenneth H Rosen |
# Lecture 21: Division and number bases
• Notation
• , quot(a, b), rem(a, b)
• bad (but common) notation: a/b, a, amodb
• Euclidean division
• statement, existence, uniqueness
• Base b representation
• interpreting a string of digits in base b
• converting to base b
## Notes on Notation
• denotes the set of all integers: ℤ = {…, − 2, −1, 0, 1, 2, …}.
• For this section of the course, unless stated otherwise, variables will be elements of . I will typically use a, b, c and n, m for integers, and will reserve variables like x and y for other types of things.
• Don't use division when working with the integers, even if you believe that things divide evenly. Find a way to write what you are saying using only multiplication and addition. It will make your life easier.
• quot(a, b) and rem(a, b) denote the quotient and remainder of a by b, as defined by the Euclidean Division Algorithm below.
• Many programming languages use a/b to denote quot(a, b). This is a bad idea and will lead to confusion. Use the same notation as in the division algorithm below.
• Many languages use a to refer to rem(a, b). This is reasonable notation, except that every programming language interprets this symbol differently when given negative numbers.
• Many books use write a mod b to refer to rem(a, b). This notation leads to massive confusion when we discuss modular arithmetic. Don't use it.
## Euclidean division
Claim (the Euclidean Division Algorithm): For any a, and any b ≠ 0, there exists integers q and r with 0 ≤ r < b and a = qb + r. Moreover, q and r are unique.
Notation: q is called the quotient of a by b, and r is called the remainder. They are written quot(a, b) and rem(a, b) respectively.
Proof (existence): We prove this only for a ≥ 0 and b > 0. You can prove the negative cases by using the existence in the positive cases.
Proof is by induction on a. Let P(a) be the statement that for all b > 0, there exists q, r as in the statement above.
To prove P(0), choose q = r = 0.
To prove P(a + 1), assume P(a). Then we know a = qb + r for some q and r. Then a + 1 = qb + (r′+1). If r′<b − 1 then r′+1 < b, so we can choose q = q and r = r′+1, which gives a = qb + r.
However, if r′=b − 1, then this choice of r doesn't satisfy the requirements of the theorem. In this case, though, we have a + 1 = qb + b = (q′+1)b + 0. Thus choosing q = q′+1 and r = 0 yields a = qb + r as required.
Proof (uniqueness): To show uniqueness, we must show that if a = qb + r and a = qb + r (with both r and r in the range [0, b)), then q = q and r = r.
Suppose that qb + r = qb + r. Then we have r − r′=(q′−q)b (). Now, we know 0 ≤ r < b and b < −r′≤0. Adding these equations yields b < r − r′<b. But we also know that r − r is a multiple of b by (), so r − r could be (for example) −2b, b, 0, b, 2b, etc. But the only multiple of b between b and b is 0, so r − r must be 0. Thus r = r.
Plugging this in to equation (*), we see that (q′−q)b = 0. Since b > 0, we have that q′−q = 0, so q = q, as required.
## Working in base b
Base b representation is a way to write numbers using the digits {0, 1, …, (b − 1)}.
Common bases:
• you use base 10 (decimal) every day (digits are {0, 1, ..., 9})
• base 2 (binary) uses digits {0, 1}. It is convenient for digital logic, a digit (called a bit) can be represented using a single wire: the wire has high voltage for 1, low for 0. Binary numbers are often designated by a trailing b: for example 1101b.
• base 16 (hexadecimal) uses the digits {0, 1, 2, ..., 9, A, B, C, D, E, F}. it is useful becausee a single digit can be represented using 4 bits. Hex numbers are often written with a prefix of "0x": for example 0xFC39.
• base 8 (octal) uses the digits {0, 1, 2, ..., 7}, and is occasionally used when 3-bit numbers are useful.
A string of digits in base b, written (anan − 1...a3a2a1a0)b, represents the number a0b0 + a1b1 + a2b2 + ⋯ + anbn.
### Arithmetic in base b
There are two ways to work with numbers in base b. The hard way: convert to base 10, apply the algorithms you already know for addition and multiplication, convert back.
The easy way: use the algorithms you already know for addition, multiplication, division, but remember that (10)b stands for b and not 10.
We did examples with long addition. Long multiplication and division work the same way.
## Writing a number in base b
Theorem: for any a and b ≥ 2, you can write a in base b. That is, there are digits d0, d1, ..., dn such that (dndn − 1...d2d1d0)b = a.
Note: as with the division algorithm, the proof of this theorem contains the algorithm used to construct the base-b representation.
Proof: by strong induction on a. In the base case, we can choose di = 0. Then since b > 1, b > d0, and a = d0 = (d0)b.
For the inductive step, assume that any number k < a can be written in base b. We wish to write a in base b. To do so, use Euclidean division to write a = qb + r. Let d0 = r. By the inductive hypothesis, we can write q in base b: q = (dldl − 1′...d2d1d0′) (we have to check that q < a, but this is true).
It turns out that these digits of q are also the higher digits of a. That is, d1 = d0 and d2 = d1 and so on. To check this, we know:
$$q = \sum_{i=0}^l d_i'b^i$$
We also know that a = qb + r, so
$$a = qb + r = b(\sum_{i=0}^l d_i'b^i) + r = d_0 + \sum_{i=0}^l d_i'b^{i+1} = d_0 + \sum_{i=0}^l d_{i+1}b^{i+1}$$
By changing i to j − 1 in the summation this becomes
$$a = d_0 + \sum_{j=1}^{l+1} d_jb^j = \sum_{j=0}^{l+1} d_jb^j$$
which shows that (di) is the base b representation of n.
Note: we stated (but did not prove) that the base b representation is unique (just as quotient and remainder are), this is a good exercise. It follows directly from the uniqueness of the quotient and remainder. |
# How do I calculate 1/3 of a number?
## How do I calculate 1/3 of a number?
Explanation:
1. Let the number be x.
2. To find one-third of a number, divide the number by 3 ,
3. x÷3.
What is 1/3 in a calculator?
1/3 = 13 ≅ 0.3333333 Spelled result in words is one third.
### What is 1/3 as a fraction?
Decimal and Fraction Conversion Chart
Fraction Equivalent Fractions
1/3 2/6 4/12
2/3 4/6 8/12
1/4 2/8 4/16
3/4 6/8 12/16
How do you find 2/3 of a fraction?
Change two-thirds to a decimal and then multiply the decimal and your number. To convert 2/3 to decimal, divide the numerator by the denominator: 2 / 3 = 0.66666 7, which you can round to 0.67. For example, to find 2/3 of 21: 0.67 * 21 = 14.07.
#### What is a third in numbers?
Toggle text. One third is one part of three equal parts. When you split an object or number into thirds, you divide it by three.
What is the ratio of 3 4?
The simplified or reduced ratio “3 to 4” tells us only that, for every three men, there are four women. The simplified ratio also tells us that, in any representative set of seven people (3 + 4 = 7) from this group, three will be men. In other words, the men comprise 73 of the people in the group.
## What is a one third?
One third is one part of three equal parts. When you split an object or number into thirds, you divide it by three. Math›What is a fraction?
What is 3/4 of a whole?
You can write it as a decimal: 0.75.
### How much is two third cups?
One half cup plus two tablespoons plus two teaspoons equals two thirds of a cup. Or, if you have the patience, ten tablespoons plus two teaspoons is also two thirds of a cup.
What is 3 2 in a whole number?
So for 3/2, the whole number is 1.
#### How many inches are in three quarters of a foot?
Three quarters of a foot can be worked out in inches by dividing a foot into quarters. There are 12 inches in a foot, and twelve divided by four is three (12 / 4 = 3). That gives the amount of inches in a quarter of a foot, but we want three quarters.
How much is a third of 12 0?
A third of 12 is 4. A third of 12 0 is 4 0. A third of 1,2 00 is 4 00. A third of 12 0,000 is 4 0,000. It is a 4 followed by four 0 ‘s. Example 2. How much is a third of \$1. 20? Solution . Ignore the decimal point. A third of 120 is 40. Now replace the decimal point. A third of \$1. 20 is \$. 40.
## How do you work out three quarters of a foot?
Three quarters of a foot can be worked out in inches by dividing a foot into quarters. There are 12 inches in a foot, and twelve divided by four is three (12 / 4 = 3). That gives the amount of inches in a quarter of a foot, but we want three quarters. So we multiply the quarter we have found by three (3 inches x 3).
How to convert one third or one third (3) (fractions)?
Really. This is a conversion chart for one third or . (3) (Fractions). To switch the unit simply find the one you want on the page and click it. You can also go to the universal conversion page. Enter the value you want to convert (one third or . (3)). Then click the Convert Me button. |
7.2 Trigonometric Integrals
# 7.2 Trigonometric Integrals
## 7.2 Trigonometric Integrals
- - - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - - -
##### Presentation Transcript
1. TECHNIQUES OF INTEGRATION 7.2Trigonometric Integrals • In this section, we will learn: • How to use trigonometric identities to integrate • certain combinations of trigonometric functions.
2. TRIGONOMETRIC INTEGRALS • We start with powers of sine and cosine.
3. SINE AND COSINE INTEGRALS Example 1 • Evaluate ∫cos3x dx • Simply substituting u = cos x is not helpful, since then du = –sin x dx. • In order to integrate powers of cosine, we would need an extra sin x factor. • Similarly, a power of sine would require an extra cos x factor.
4. SINE AND COSINE INTEGRALS Example 1 • Thus, here we can separate one cosine factor and convert the remaining cos2x factor to an expression involving sine using the identity sin2x + cos2x = 1: • cos3x = cos2x .cosx = (1 - sin2x) cosx
5. SINE AND COSINE INTEGRALS Example 1 • We can then evaluate the integral by substituting u = sin x. So, du = cos x dx and
6. SINE AND COSINE INTEGRALS • In general, we try to write an integrand involving powers of sine and cosine in a form where we have only one sine factor. • The remainder of the expression can be in terms of cosine.
7. SINE AND COSINE INTEGRALS • We could also try only one cosine factor. • The remainder of the expression can be in terms of sine.
8. SINE AND COSINE INTEGRALS • The identity sin2x + cos2x = 1 • enables us to convert back and forth between even powers of sine and cosine.
9. SINE AND COSINE INTEGRALS Example 2 • Find ∫sin5x cos2x dx • We could convert cos2x to 1 – sin2x. • However, we would be left with an expression in terms of sin x with no extra cos x factor.
10. SINE AND COSINE INTEGRALS Example 2 • Instead, we separate a single sine factor and rewrite the remaining sin4x factor in terms of cos x. • So, we have:
11. SINE AND COSINE INTEGRALS Example 2 • Substituting u=cos x, we have du =-sin x dx. So,
12. SINE AND COSINE INTEGRALS • The figure shows the graphs of the integrand sin5x cos2x in Example 2 and its indefinite integral (with C = 0).
13. SINE AND COSINE INTEGRALS • In the preceding examples, an odd power of sine or cosine enabled us to separate a single factor and convert the remaining even power. • If the integrand contains even powers of both sine and cosine, this strategy fails.
14. SINE AND COSINE INTEGRALS • In that case, we can take advantage of the following half-angle identities:
15. SINE AND COSINE INTEGRALS Example 3 • Evaluate • If we write sin2x = 1 - cos2x, the integral is no simpler to evaluate.
16. SINE AND COSINE INTEGRALS Example 3 • However, using the half-angle formula for sin2x, we have:
17. SINE AND COSINE INTEGRALS Example 3 • Notice that we mentally made the substitution • u = 2x when integrating cos 2x. • Another method for evaluating this integral was given in Exercise 43 in Section 7.1
18. SINE AND COSINE INTEGRALS Example 4 • Find • We could evaluate this integral using the reduction formula for ∫sinnx dx (Equation 7 in Section 7.1) together with Example 3.
19. SINE AND COSINE INTEGRALS Example 4 • However, a better method is to write and use a half-angle formula:
20. SINE AND COSINE INTEGRALS Example 4 • As cos2 2x occurs, we must use another half-angle formula:
21. SINE AND COSINE INTEGRALS Example 4 • This gives:
22. SINE AND COSINE INTEGRALS • To summarize, we list guidelines to follow when evaluating integrals of the form • where m ≥ 0 and n≥ 0 are integers.
23. STRATEGY A • If the power of cosine is odd (n = 2k + 1), save one cosine factor. • Use cos2x = 1 - sin2x to express the remaining factors in terms of sine: • Then, substitute u = sin x.
24. STRATEGY B • If the power of sine is odd (m = 2k + 1), save one sine factor. • Use sin2x= 1 - cos2x to express the remaining factors in terms of cosine: • Then, substitute u = cos x.
25. STRATEGIES • Note that, if the powers of both sine and cosine are odd, either (A) or (B) can be used.
26. STRATEGY C • If the powers of both sine and cosine are even, use the half-angle identities • Sometimes, it is helpful to use the identity
27. TANGENT & SECANT INTEGRALS • We can use a similar strategy to evaluate integrals of the form
28. TANGENT & SECANT INTEGRALS • As (d/dx)tan x = sec2x, we can separate a sec2x factor. • Then, we convert the remaining (even) power of secant to an expression involving tangent using the identity sec2x = 1 + tan2x.
29. TANGENT & SECANT INTEGRALS • Alternately, as (d/dx) sec x = sec x tan x, we can separate a sec x tan x factor and convert the remaining (even) power of tangent to secant.
30. TANGENT & SECANT INTEGRALS Example 5 • Evaluate∫tan6x sec4x dx • If we separate one sec2x factor, we can express the remaining sec2x factor in terms of tangent using the identity sec2x = 1 + tan2x. • Then, we can evaluate the integral by substituting u = tan x so that du = sec2xdx.
31. TANGENT & SECANT INTEGRALS Example 5 • We have:
32. TANGENT & SECANT INTEGRALS Example 6 • Find∫ tan5 θsec7θ dθ • If we separate a sec2θfactor, as in the preceding example, we are left with a sec5θfactor. • This is not easily converted to tangent.
33. TANGENT & SECANT INTEGRALS Example 6 • However, if we separate a sec θ tan θ factor, we can convert the remaining power of tangent to an expression involving only secant. • We can use the identity tan2θ = sec2θ – 1.
34. TANGENT & SECANT INTEGRALS Example 6 • We then evaluate the integral by substituting u = sec θ, so du = sec θtan θdθ:
35. TANGENT & SECANT INTEGRALS • The preceding examples demonstrate strategies for evaluating integrals in the form ∫tanmx secnx for two cases—which we summarize here.
36. STRATEGY A • If the power of secant is even (n = 2k, k≥ 2) save sec2x. • Then, use tan2x = 1 + sec2x to express the remaining factors in terms of tan x: • Then, substitute u = tan x.
37. STRATEGY B • If the power of tangent is odd (m = 2k + 1), save sec x tan x. • Then, use tan2x = sec2x – 1 to express the remaining factors in terms of sec x: • Then, substitute u = sec x.
38. OTHER INTEGRALS • For other cases, the guidelines are not as clear-cut. • We may need to use: • Identities • Integration by parts • A little ingenuity
39. TANGENT & SECANT INTEGRALS • We will need to be able to integrate tan x by using a substitution,
40. TANGENT & SECANT INTEGRALS Formula 1 • We will also need the indefinite integral of secant: • We could verify Formula 1 by differentiating the right side, or as follows.
41. TANGENT & SECANT INTEGRALS • First, we multiply numerator and denominator by sec x + tan x:
42. TANGENT & SECANT INTEGRALS • If we substitute u = sec x + tan x, then • du = (sec x tan x + sec2x) dx. • The integral becomes: ∫ (1/u) du = ln |u| +C • Thus, we have:
43. TANGENT & SECANT INTEGRALS Example 7 • Find • Here, only tan x occurs. • So, we rewrite a tan2x factor in terms of sec2x.
44. TANGENT & SECANT INTEGRALS Example 7 • Hence, we use tan2x - sec2x = 1. • In the first integral, we mentally substituted u = tan x so that du = sec2x dx.
45. TANGENT & SECANT INTEGRALS • If an even power of tangent appears with an odd power of secant, it is helpful to express the integrand completely in terms of sec x. • Powers of sec x may require integration by parts, as shown in the following example.
46. TANGENT & SECANT INTEGRALS Example 8 • Find • Here, we integrate by parts with
47. TANGENT & SECANT INTEGRALS Example 8 • Then,
48. TANGENT & SECANT INTEGRALS Example 8 • Using Formula 1 and solving for the required integral, we get:
49. TANGENT & SECANT INTEGRALS • Integrals such as the one in the example may seem very special. • However, they occur frequently in applications of integration.
50. COTANGENT & COSECANT INTEGRALS • Integrals of the form∫cotmx cscnx dx can be found by similar methods. • We have to make use of the identity 1 + cot2x = csc2x |
# The equation of the tangent line to the curve at the point.
### Single Variable Calculus: Concepts...
4th Edition
James Stewart
Publisher: Cengage Learning
ISBN: 9781337687805
### Single Variable Calculus: Concepts...
4th Edition
James Stewart
Publisher: Cengage Learning
ISBN: 9781337687805
#### Solutions
Chapter 3.4, Problem 45E
(a)
To determine
## To find: The equation of the tangent line to the curve at the point.
Expert Solution
The equation of the tangent line to the curve y=2(1+ex) at (0,1) is y=x2+1.
### Explanation of Solution
Given:
The function is y=2(1+ex).
Derivative Rule: Quotient Rule
If f(x). and g(x) are both differentiable function, then
ddx[f(x)g(x)]=g(x)ddx[f(x)]f(x)ddx[g(x)][g(x)]2 (1)
Formula used:
The equation of the tangent line at (x1,y1) is, yy1=m(xx1) (2)
where, m is the slope of the tangent line at (x1,y1) and m=dydx|x=x1.
Calculation:
The derivative of y is dydx, which is obtained as follows,
dydx=ddx(y)=ddx(2(1+ex))
Apply the quotient rule as shown in equation (1),
ddx[2(1+ex)]=(1+ex)ddx[2]2ddx[(1+ex)](1+ex)2=(1+ex)[0]2(ddx(1)+ddx(ex))(1+ex)2=02(0+(ex))(1+ex)2=2(ex)(1+ex)2
Therefore, the derivative of y=2(1+ex) is dydx=2(ex)(1+ex)2_.
The slope of the tangent line at (0,1) is computed as follows,
m=dydx|x=02(e0)(1+e0)2 =2(1)(2)2 [Qe0=1]=12
Thus, the slope of the tangent line at (0,1) is m=12.
Substitute (0,1) for (x1,y1) and m=12 in equation (1),
(y1)=12(x0)y1=x2y=x2+1
Therefore, the equation of the tangent line to the curve y=2(1+ex) at (0,1) is y=x2+1.
(b)
To determine
Expert Solution
### Explanation of Solution
Given:
The equation of the curve is y=2(1+ex).
The equation of the tangent line is y=x2+1.
Graph:
Use the online graphing calculator to draw the graph of the functions as shown below in Figure 1.
From the Figure 1, it is observed that the equation of the tangent line touches on the curve y=2(1+ex) at the point (0,1).
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# If $\alpha$ and $\beta$ are the zeros of the quadratic polynomial $f\left( x \right) = {x^2} - 5x + 4$, find the value of $\dfrac{1}{\alpha } + \dfrac{1}{\beta } - 2\alpha \beta$.
Last updated date: 25th Jul 2024
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Hint: Zeros of the quadratic equation are the values of the dependent variable for which the quadratic expression becomes 0.
To find the value of zeros, put f(x)=0.
$\Rightarrow$ ${x^2} - 5x + 4 = 0$
$\Rightarrow$ ${x^2} - 4x - x + 4 = 0$
$\Rightarrow$ $x(x - 4) -1 (x - 4) = 0$
$\Rightarrow \left( {x - 1} \right)\left( {x - 4} \right) = 0$
Zeros of the quadratic polynomial are
$\Rightarrow \alpha = 1, \beta = 4$
Now,
$\Rightarrow \dfrac{1}{\alpha } + \dfrac{1}{\beta } - 2\alpha \beta = \dfrac{1}{1} + \dfrac{1}{4} - 2 \times 1 \times 4$
$\Rightarrow \dfrac{5}{4} - 8$
$\Rightarrow - \dfrac{{27}}{4}$
Note: Zeros is the intersection of the polynomial and the axis, if the polynomial is in x, then zeros is the intersection of the polynomial with x-axis. The roots can also be found using the quadratic formula. |
# How do you simplify sqrt84 * sqrt28?
Apr 30, 2016
$\sqrt{84} \cdot \sqrt{28} = \sqrt{{28}^{2} \cdot 3} = 28 \sqrt{3}$
#### Explanation:
Here's a factor tree for $28$:
$\textcolor{w h i t e}{0000} 28$
$\textcolor{w h i t e}{000} \text{/"color(white)(00)"\}$
$\textcolor{w h i t e}{00} 2 \textcolor{w h i t e}{000} 14$
$\textcolor{w h i t e}{00000} \text{/"color(white)(00)"\}$
$\textcolor{w h i t e}{0000} 2 \textcolor{w h i t e}{0000} 7$
So: $28 = {2}^{2} \cdot 7$
$\textcolor{w h i t e}{}$
Here's a factor tree for $84$:
$\textcolor{w h i t e}{0000} 84$
$\textcolor{w h i t e}{000} \text{/"color(white)(00)"\}$
$\textcolor{w h i t e}{00} 2 \textcolor{w h i t e}{000} 42$
$\textcolor{w h i t e}{00000} \text{/"color(white)(00)"\}$
$\textcolor{w h i t e}{0000} 2 \textcolor{w h i t e}{000} 21$
$\textcolor{w h i t e}{0000000} \text{/"color(white)(00)"\}$
$\textcolor{w h i t e}{000000} 3 \textcolor{w h i t e}{0000} 7$
So: $84 = {2}^{2} \cdot 3 \cdot 7 = 28 \cdot 3$
$\textcolor{w h i t e}{}$
So:
$\sqrt{84} \cdot \sqrt{28} = \sqrt{84 \cdot 28} = \sqrt{{28}^{2} \cdot 3} = 28 \sqrt{3}$ |
18 percent of 345
Here we will show you how to calculate eighteen percent of three hundred forty-five. Before we continue, note that 18 percent of 345 is the same as 18% of 345. We will write it both ways throughout this tutorial to remind you that it is the same.
18 percent means that for each 100, there are 18 of something. This page will teach you three different methods you can use to calculate 18 percent of 345.
We think that illustrating multiple ways of calculating 18 percent of 345 will give you a comprehensive understanding of what 18% of 345 means, and provide you with percent knowledge that you can use to calculate any percentage in the future.
To solidify your understanding of 18 percent of 345 even further, we have also created a pie chart showing 18% of 345. On top of that, we will explain and calculate "What is not 18 percent of 345?"
Calculate 18 percent of 345 using a formula
This is the most common method to calculate 18% of 345. 345 is the Whole, 18 is the Percent, and the Part is what we are calculating. Below is the math and answer to "What is 18% of 345?" using the percent formula.
(Whole × Percent)/100 = Part
(345 × 18)/100 = 62.1
18% of 345 = 62.1
Get 18 percent of 345 with a percent decimal number
You can convert any percent, such as 18.00%, to 18 percent as a decimal by dividing the percent by one hundred. Therefore, 18% as a decimal is 0.18. Here is how to calculate 18 percent of 345 with percent as a decimal.
Whole × Percent as a Decimal = Part
345 × 0.18 = 62.1
18% of 345 = 62.1
Get 18 percent of 345 with a fraction function
This is our favorite method of calculating 18% of 345 because it best illustrates what 18 percent of 345 really means. The facts are that it is 18 per 100 and we want to find parts per 345. Here is how to illustrate and show you the answer using a function with fractions.
Part 345
=
18 100
Part = 62.1
18% of 345 = 62.1
Note: To solve the equation above, we first multiplied both sides by 345 and then divided the left side to get the answer.
18 percent of 345 illustrated
Below is a pie chart illustrating 18 percent of 345. The pie contains 345 parts, and the blue part of the pie is 62.1 parts or 18 percent of 345.
Note that it does not matter what the parts are. It could be 18 percent of 345 dollars, 18 percent of 345 people, and so on. The pie chart of 18% of 345 will look the same regardless what it is.
What is not 18 percent of 345?
What is not 18 percent of 345? In other words, what is the red part of our pie above? We know that the total is 100 percent, so to calculate "What is not 18%?" you deduct 18% from 100% and then take that percent from 345:
100% - 18% = 82%
(345 × 82)/100 = 282.9
Another way of calculating the red part is to subtract 62.1 from 345.
345 - 62.1 = 282.9
That is the end of our tutorial folks. We hope we accomplished our goal of making you a percent expert - at least when it comes to calculating 18 percent of 345.
Percent of a Number
Go here if you need to calculate the percent of a different number.
18 percent of 346
Here is the next percent tutorial on our list that may be of interest. |
# Determine the slope of a line given a table of values, a graph, two points on the line, and an equation written in various forms, including y = mx + b, ax + by = c, and y - y1 = m(x - x1)
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# Integers - Mathematics Form 1 Notes
## Introduction
### The Number Line
• Integers are whole numbers, negative whole numbers and zero.
• Integers are always represented on the number line at equal intervals which are equal to one unit.
## Operations on Integers
• Addition of integers can be represented on a number line .
• For example, to add +3 to 0 , we begin at 0 and move 3 units to the right as shown below in red to get +3,
• Also to add + 4 to +3 we move 4 units to the right as shown in blue to get +7.
• To add -3 to zero we move 3 units to the left as shown in red below to get -3 while to add -2 to -3 we move 2 steps to the left as shown in blue to get -5.
Note;
• When adding positive numbers we move to the right.
• When dealing with negative we move to the left.
### Subtraction of Integers.
Example
(+7) – (0) = (+7)
To subtract +7 from 0 ,we find a number n which when added to get 0 we get +7 and in this case n = +7 as shown above in red.
Example
(+2) – (+7) = (-5)
Start at +7 and move to +2. 5 steps will be made towards the left. The answer is therefore -5.
Example
-3 – (+6) = -9
|__|_←|__|__|__|__|__| ||__|__|__|__|__|
-4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10
We start at +6 and moves to -3. 9 steps to the left, the answer is -9.
Note:
• In general positives signs can be ignored when writing positive numbers i.e. +2 can be written as 2 but negative signs cannot be ignored when writing negative numbers -4 can only be written s -4.
4 – (+3) = 4 -3
= 1
-3- (+6) =3 – 6
= -3
• Positive integers are also referred to as natural numbers. The result of subtracting the negative of a number is the same as adding that number.
2 – (- 4) = 2 + 4
= 6
(-5) – (- 1 ) = -5 + 2
= -3
• In mathematics it is assumed that that the number with no sign before it has appositive sign.
### Multiplication
- In general
1. (a negative number) x (appositive number ) = (a negative number)
2. (a positive number) x (a negative number ) = (a negative number)
3. (a negative number) x (a negative number ) = (a positive number)
Examples
-6 x 5 = -30
7 x -4 = - 28
-3 x -3 = 9
-2 x -9 = 18
### Division
- Division is the inverse of multiplication. In general
1. (a positive number ) ÷ (a positive number ) = (a positive number)
2. (a positive number ) ÷ (a negative number ) = (a negative number)
3. (a negative number ) ÷ (a negative number ) = (a positive number)
4. (a negative number ) ÷ (appositive number ) = (a negative number)
- For multiplication and division of integer:
• Two like signs gives positive sign.
• Two unlike signs gives negative sign
• Multiplication by zero is always zero and division by zero is always zero.
### Order of Operations
• BODMAS is always used to show as the order of operations.
B – Bracket first.
O – Of is second.
D – Division is third.
M – Multiplication is fourth.
S – Subtraction is considered last.
Example
6 x 3 – 4 ÷ 2 + 5 + (2-1) =
Solution
Use BODMAS
(2 – 1 ) = 1 we solve brackets first
(4÷ 2) = 2 we then solve division
(6 x 3) = 1 8 next is multiplication
Bring them together
18 – 2 +5 +1 = 22 we solve addition first and lastly subtraction
18 + 6 – 2 = 22
## Past KCSE Questions on the Topic
1. The sum of two numbers exceeds their product by one. Their difference is equal to their product less five.Find the two numbers. (3mks)
2. 3x – 1 > -4
2x + 1 ≤ 7
3. Evaluate
-12 ÷ (-3) x 4 – (-15)
-5 x 6 ÷ 2 + (-5)
4. Without using a calculator/mathematical tables, evaluate leaving your answer as a simple fraction
(-4)(-2) + (-12) ÷ (+3) + -20 + (+4) + -6)
-9 – (15) 46- (8+2)-3
5. Evaluate -8 ÷ 2 + 12 x 9 – 4 x 6
56 ÷7 x 2
6. Evaluate without using mathematical tables or the calculator
1.9 x 0.032
20 x 0.0038
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# Exponential Growth
## Functions with x as an exponent
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Exponential Growth Function
A population of 10 mice grows at a rate of 300% every month. How many mice are in the population after six months?
### Guidance
An exponential function has the variable in the exponent of the expression. All exponential functions have the form: f(x)=abxh+k\begin{align*}f(x)=a \cdot b^{x-h}+k\end{align*}, where \begin{align*}h\end{align*} and \begin{align*}k\end{align*} move the function in the \begin{align*}x\end{align*} and \begin{align*}y\end{align*} directions respectively, much like the other functions we have seen in this text. \begin{align*}b\end{align*} is the base and \begin{align*}a\end{align*} changes how quickly or slowly the function grows. Let’s take a look at the parent graph, \begin{align*}y=2^x\end{align*}.
#### Example A
Graph \begin{align*}y=2^x\end{align*}. Find the \begin{align*}y\end{align*}-intercept.
Solution: Let’s start by making a table. Include some positive and negative values for \begin{align*}x\end{align*} and zero.
\begin{align*}x\end{align*} \begin{align*}2^x\end{align*} \begin{align*}y\end{align*}
3 \begin{align*}2^3\end{align*} 8
2 \begin{align*}2^2\end{align*} 4
1 \begin{align*}2^1\end{align*} 2
0 \begin{align*}2^0\end{align*} 1
-1 \begin{align*}2^{-1}\end{align*} \begin{align*}\frac{1}{2}\end{align*}
-2 \begin{align*}2^{-2}\end{align*} \begin{align*}\frac{1}{4}\end{align*}
-3 \begin{align*}2^{-3}\end{align*} \begin{align*}\frac{1}{8}\end{align*}
This is the typical shape of an exponential growth function. The function grows “exponentially fast”. Meaning, in this case, the function grows in powers of 2. For an exponential function to be a growth function, \begin{align*}a > 0\end{align*} and \begin{align*}b > 1\end{align*} and \begin{align*}h\end{align*} and \begin{align*}k\end{align*} are both zero \begin{align*}(y=ab^x)\end{align*}. From the table, we see that the \begin{align*}y\end{align*}-intercept is (0, 1).
Notice that the function gets very, very close to the \begin{align*}x\end{align*}-axis, but never touches or passes through it. Even if we chose \begin{align*}x=-50, \ y\end{align*} would be \begin{align*}2^{-50}=\frac{1}{2^{50}}\end{align*}, which is still not zero, but very close. In fact, the function will never reach zero, even though it will get smaller and smaller. Therefore, this function approaches the line \begin{align*}y=0\end{align*}, but will never touch or pass through it. This type of boundary line is called an asymptote. In the case with all exponential functions, there will be a horizontal asymptote. If \begin{align*}k=0\end{align*}, then the asymptote will be \begin{align*}y=0\end{align*}.
#### Example B
Graph \begin{align*}y=3^{x-2}+1\end{align*}. Find the \begin{align*}y\end{align*}-intercept, asymptote, domain and range.
Solution: This is not considered a growth function because \begin{align*}h\end{align*} and \begin{align*}k\end{align*} are not zero. To graph something like this (without a calculator), start by graphing \begin{align*}y=3^x\end{align*} and then shift it \begin{align*}h\end{align*} units in the \begin{align*}x\end{align*}-direction and \begin{align*}k\end{align*} units in the \begin{align*}y\end{align*}-direction.
Notice that the point (0, 1) from \begin{align*}y=3^x\end{align*} gets shifted to the right 2 units and up one unit and is (2, 2) in the translated function, \begin{align*}y=3^{x-2}+1\end{align*}. Therefore, the asymptote is \begin{align*}y=1\end{align*}. To find the \begin{align*}y\end{align*}-intercept, plug in \begin{align*}x=0\end{align*}.
The domain of all exponential functions is all real numbers. The range will be everything greater than the asymptote. In this example, the range is \begin{align*}y > 1\end{align*}.
#### Example C
Graph the function \begin{align*}y= -\frac{1}{2} \cdot 4^x\end{align*}. Determine if it is an exponential growth function.
Solution: In this example, we will outline how to use the graphing calculator to graph an exponential function. First, clear out anything in Y=. Next, input the function into Y1= -(1/2)4^X and press GRAPH. Adjust your window accordingly.
This is not an exponential growth function, because it does not grow in a positive direction. By looking at the definition of a growth function, \begin{align*}a>0\end{align*}, and it is not here.
Intro Problem Revisit
This is an example of exponential growth, so we can use the exponential form \begin{align*}f(x)=a \cdot b^{x-h}+k\end{align*}. In this case, a = 10, the starting population; b = 300% or 3, the rate of growth; x-h = 6 the number of months, and k = 0.
Therefore, the mouse population after six months is 7,290.
### Guided Practice
Graph the following exponential functions. Determine if they are growth functions. Then, find the \begin{align*}y\end{align*}-intercept, asymptote, domain and range. Use an appropriate window.
1. \begin{align*}y=3^{x-4}-2\end{align*}
2. \begin{align*}f(x)=(-2)^{x+5}\end{align*}
3. \begin{align*}f(x)=5^x\end{align*}
4. Abigail is in a singles tennis tournament. She finds out that there are eight rounds until the final match. If the tournament is single elimination, how many games will be played? How many competitors are in the tournament?
1. This is not a growth function because \begin{align*}h\end{align*} and \begin{align*}k\end{align*} are not zero. The \begin{align*}y\end{align*}-intercept is \begin{align*}y=3^{0-4}-2=\frac{1}{81}-2=-1\frac{80}{81}\end{align*}, the asymptote is at \begin{align*}y=-2\end{align*}, the domain is all real numbers and the range is \begin{align*}y>-2\end{align*}.
2. This is not a growth function because \begin{align*}h\end{align*} is not zero. The \begin{align*}y\end{align*}-intercept is \begin{align*}y=(-2)^{0+5}=(-2)^5=-32\end{align*}, the asymptote is at \begin{align*}y=0\end{align*}, the domain is all real numbers and the range is \begin{align*}y>0\end{align*}.
3. This is a growth function. The \begin{align*}y\end{align*}-intercept is \begin{align*}y=5^\circ =1\end{align*}, the asymptote is at \begin{align*}y=0\end{align*}, the domain is all real numbers and the range is \begin{align*}y>0\end{align*}.
4. If there are eight rounds to single’s games, there are will be \begin{align*}2^8=256\end{align*} competitors. In the first round, there will be 128 matches, then 64 matches, followed by 32 matches, then 16 matches, 8, 4, 2, and finally the championship game. Adding all these all together, there will be \begin{align*}128+64+32+16+8+4+2+1\end{align*} or 255 total matches.
### Explore More
Graph the following exponential functions. Find the \begin{align*}y\end{align*}-intercept, the equation of the asymptote and the domain and range for each function.
1. \begin{align*}y=4^x\end{align*}
2. \begin{align*}y=(-1)(5)^x\end{align*}
3. \begin{align*}y=3^x-2\end{align*}
4. \begin{align*}y=2^x+1\end{align*}
5. \begin{align*}y=6^{x+3}\end{align*}
6. \begin{align*}y= -\frac{1}{4}(2)^x+3\end{align*}
7. \begin{align*}y=7^{x+3}-5\end{align*}
8. \begin{align*}y=-(3)^{x-4}+2\end{align*}
9. \begin{align*}y=3(2)^{x+1}-5\end{align*}
10. What is the y-intercept of \begin{align*}y=a^x\end{align*}? Why is that?
11. What is the range of the function \begin{align*}y=a^{x-h}+k\end{align*}?
12. March Madness is a single-game elimination tournament of 64 college basketball teams. How many games will be played until there is a champion? Include the championship game.
13. In 2012, the tournament added 4 teams to make it a field of 68 and there are 4 "play-in" games at the beginning of the tournament. How many games are played now?
14. An investment grows according the function \begin{align*}A=P(1.05)^t\end{align*} where \begin{align*}P\end{align*} represents the initial investment, \begin{align*}A\end{align*} represents the value of the investment and \begin{align*}t\end{align*} represents the number of years of investment. If \$10,000 was the initial investment, how much would the value of the investment be after 10 years, to the nearest dollar?
15. How much would the value of the investment be after 20 years, to the nearest dollar?
### Vocabulary Language: English
Asymptotes
Asymptotes
An asymptote is a line on the graph of a function representing a value toward which the function may approach, but does not reach (with certain exceptions).
Exponential Function
Exponential Function
An exponential function is a function whose variable is in the exponent. The general form is $y=a \cdot b^{x-h}+k$.
Exponential growth
Exponential growth
Exponential growth occurs when a quantity increases by the same proportion in each given time period.
Exponential Growth Function
Exponential Growth Function
An exponential growth function is a specific type of exponential function that has the form $y=ab^x$, where $h=k=0, a>0,$ and $b>1$.
Model
Model
A model is a mathematical expression or function used to describe a physical item or situation. |
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# Class 8 RD Sharma Solutions – Chapter 3 Squares and Square Roots – Exercise 3.4 | Set 2
### Question 11. The area of a square field is 5184 m2. A rectangular field, whose length is twice its breadth has its perimeter equal to the perimeter of the square field. Find the area of the rectangular field.
Solution:
Let the side of square field as x
x2 = 5184 m2
x = √5184m
x = 2 × 2 ×2 × 9
= 72 m
Perimeter of square = 4x
= 4(72)
= 288 m
Perimeter of rectangle = 2 (l + b) = perimeter of the square field
= 288 m
l = 2b
2 (2b + b) = 288
2(3b) = 288
6b = 288
b = 288/6 (Transposing 6)
b = 48m
l = 2 × 48
= 96m
Area of rectangle = l × b
Area of rectangle = 96 × 48 m2
= 4608 m2
### Question 12. Find the least square number, exactly divisible by each one of the numbers:
(i) 6, 9, 15 and 20
Solution:
L.C.M of 6, 9, 15, 20 is 180
Prime factorization of 180 = 22 × 32 × 5 (Pairing of 2 and 3)
5 is left out
Multiplying the number with 5
180 × 5 = 22 × 32 × 52
= 900
Therefore, 900 is the least square number divisible by 6, 9, 15 and 20
(ii) 8, 12, 15 and 20
Solution:
L.C.M of 8, 2, 15, 20 is 360
Prime factorization of 360 = 22 × 32 ×2 × 5
2 and 5 are left out
Multiplying the number with 2 × 5 = 10
360 × 10 = 22 × 32 × 52 × 22
Therefore, 3600 is the least square number divisible by 8, 12, 15 and 20
### Question 13. Find the square roots of 121 and 169 by the method of repeated subtraction.
Solution:
In repeated subtraction method, odd numbers are subtracted one by one from the previous result and number of times subtraction is carried out is the square root.
121 – 1 = 120
120 – 3 = 117
117 – 5 = 112
112 – 7 = 105
105 – 9 = 96
96 – 11 = 85
85 – 13 = 72
72 – 15 = 57
57 – 17 = 40
40 – 19 = 21
21 – 21 = 0
11 times subtraction operation is carried out
Therefore, √121 = 11
169 – 1 = 168
168 – 3 = 165
165 – 5 = 160
160 – 7 = 153
153 – 9 = 144
144 – 11 = 133
133 – 13 = 120
120 – 15 = 105
105 – 17 = 88
88 – 19 = 69
69 – 21 = 48
48 – 23 = 25
25 – 25 = 0
13 times subtraction operation is carried out
Therefore, √169 = 13
### Question 14. Write the prime factorization of the following numbers and hence find their square roots.
(i) 7744
Solution:
Prime factorization of 7744 is
7744 = 22 × 22 × 22 × 112
Therefore, the square root of 7744 is
√7744 = 2 × 2 × 2 × 11
= 88
(ii) 9604
Solution:
Prime factorization of 9604 is
9604 = 22 × 72 × 72
Therefore, the square root of 9604 is
√9604 = 2 × 7 × 7
= 98
(iii) 5929
Solution:
Prime factorization of 5929 is
5929 = 112 × 72
Therefore, the square root of 5929 is
√5929 = 11 × 7
= 77
(iv) 7056
Solution:
Prime factorization of 7056 is
7056 = 22 × 22 × 72 × 32
Therefore, the square root of 7056 is
√7056 = 2 × 2 × 7 × 3
= 84
### Question 15. The students of class VIII of a school donated Rs 2401 for PM’s National Relief Fund. Each student donated as many rupees as the number of students in the class, Find the number of students in the class.
Solution:
Let the number of students be x
Each student denoted x rupees
Total amount collected is x × x rupees = 2401
x2 = 2401
x = √2401
x = 49
Therefore, there are 49 students in the class.
### Question 16. A PT teacher wants to arrange maximum possible number of 6000 students in a field such that the number of rows is equal to the number of columns. Find the number of rows if 71 were left out after arrangement.
Solution:
Let the number of rows be x
Number of columns = x
Total number of students in the arrangement = x2
71 students are left out
Total students x2 + 71 = 6000
x2 = 5929
x = √5929
x = 77
Therefore, total number of rows are 77.
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# Tangent Equation of Circle & Point of Contact
Intersection of straight line and a circle: Let the equation of circle be x² + y² = a² and the equation of the line be y = mx + c then
When points of intersection are real and distinct: Then length of perpendicular from centre should be less the radius.
∴ $$\left| \frac{c}{\sqrt{1+{{m}^{2}}}} \right|<a$$.
When points of intersection are coincident: Line touches the circle if the length perpendicular from centre is equal to radius.
∴ $$\left| \frac{c}{\sqrt{1+{{m}^{2}}}} \right|=a$$.
When points of intersection are imaginary: Line does not intersect a circle if the length of perpendicular from centre is greater than radius of circle.
∴ $$\left| \frac{c}{\sqrt{1+{{m}^{2}}}} \right|>a$$.
Length of the intercept cut off from the line y = mx + c by the circle x² + y² = a² is $$2\sqrt{\frac{{{a}^{2}}\left( 1+{{m}^{2}} \right)-{{c}^{2}}}{1+{{m}^{2}}}}$$.
Condition of Tangency: The line y = mx + c touches the circle x² + y² = a² if the length of the intercepts is zero i.e., c = ± a √(1 + m²).
Tangent to a circle: Let P be a point on circle and let PQ be secant.
Tangent at point P is the limiting position of a secant PQ when Q tends to P along the circle. The point P is called the point of contact of the tangent.
Different forms of equation of tangents:
⇒ The equation of tangent of slope m to the circle x² + y² = a² is y = mx ± a √(1 + m²) the coordinates of the point of contact are $$\left( \pm \frac{am}{\sqrt{1+{{m}^{2}}}},\,\,\mp \frac{a}{\sqrt{1+{{m}^{2}}}} \right)$$.
⇒ The equations of tangents of slope m to the circle (x – a)² + (y – b)² = r² are given by y – b = m (x – a) ± r √(1 + m²) and the coordinates of point of contact are $$\left( a\pm \frac{mr}{\sqrt{1+{{m}^{2}}}},\,\,b\mp \frac{r}{\sqrt{1+{{m}^{2}}}} \right)$$.
⇒ The equations of the tangents of slope m to circle x² + y² + 2gx + 2fy + c = 0 is y + f = m(x + g) ± $$\sqrt{\left( {{g}^{2}}+{{f}^{2}}-c \right)\left( 1+{{m}^{2}} \right)}$$.
Point form: The equation of the tangent at the point P(x₁, y₁) to a circle x² + y² + 2gx + 2fy + c = 0 is xx₁ + yy₁ + g(x + x₁) + f(y + y₁) + c = 0.
Parametric form: The equation of the tangent to the circle x² + y² = a² at the point (a cosθ, a sinθ) is x cosθ + y sinθ = a.
The equation of tangent to the circle (x – a)² + (y – b)² = r² at the point (a + r cosθ, b + r sinθ) is (x – a) cosθ + (y – b) sinθ = r. |
## The Circumcentre of a Triangle: Exploring its Properties and Applications
Triangles are fundamental geometric shapes that have fascinated mathematicians and scientists for centuries. One intriguing aspect of triangles is their circumcentre, a point that holds significant properties and applications in various fields. In this article, we will delve into the concept of the circumcentre, explore its properties, and discuss its relevance in different contexts.
## Understanding the Circumcentre
The circumcentre of a triangle is the point where the perpendicular bisectors of the triangle’s sides intersect. It is the center of the circle that passes through all three vertices of the triangle. This point is denoted as O and is equidistant from the three vertices of the triangle.
To visualize the circumcentre, let’s consider an example. Take a triangle with vertices A, B, and C. The perpendicular bisectors of the sides AB, BC, and CA intersect at a single point, which is the circumcentre O. This point O is equidistant from A, B, and C, forming a circle that passes through all three vertices.
## Properties of the Circumcentre
The circumcentre possesses several interesting properties that make it a valuable concept in geometry. Let’s explore some of these properties:
### 1. Equidistance from Vertices
As mentioned earlier, the circumcentre is equidistant from the three vertices of the triangle. This property implies that the distances OA, OB, and OC are equal, where O is the circumcentre and A, B, and C are the vertices of the triangle.
### 2. Intersection of Perpendicular Bisectors
The circumcentre is the point of intersection of the perpendicular bisectors of the triangle’s sides. The perpendicular bisector of a side is a line that divides the side into two equal halves and is perpendicular to that side. The circumcentre is the only point where all three perpendicular bisectors intersect.
### 3. Unique Existence
Every non-degenerate triangle has a unique circumcentre. This means that for any given triangle, there is only one point that satisfies the conditions of being equidistant from the vertices and the intersection of the perpendicular bisectors.
### 4. Relationship with Orthocentre
The circumcentre and orthocentre of a triangle are related in an interesting way. The orthocentre is the point of intersection of the triangle’s altitudes, which are the perpendiculars drawn from each vertex to the opposite side. The line segment joining the circumcentre and orthocentre is called the Euler line, and it passes through the midpoint of the line segment joining the triangle’s circumcentre and centroid.
## Applications of the Circumcentre
The concept of the circumcentre finds applications in various fields, including mathematics, physics, and computer science. Let’s explore some of these applications:
### 1. Triangle Construction
The circumcentre plays a crucial role in constructing triangles. Given three points, constructing a triangle with those points as vertices involves finding the circumcentre. This construction is useful in various fields, such as architecture, engineering, and computer graphics.
### 2. Triangulation Algorithms
In computational geometry, triangulation algorithms are used to partition a given space into triangles. The circumcentre is utilized in these algorithms to determine the optimal placement of vertices and edges, ensuring the resulting triangles are well-formed and have desirable properties.
### 3. Optimal Location Determination
In certain optimization problems, determining the optimal location of a point or object is crucial. The circumcentre can be used to find the optimal location that minimizes the sum of distances to a set of points. This concept is applied in various fields, including facility location planning, transportation network design, and wireless communication network optimization.
### 4. Geometric Analysis
The circumcentre is often used in geometric analysis to study the properties and relationships of triangles. It helps in proving theorems, solving geometric problems, and understanding the behavior of triangles in different scenarios. The circumcentre’s properties provide valuable insights into the nature of triangles and their geometric properties.
## Q&A
### 1. Can a triangle have its circumcentre outside the triangle?
No, a triangle’s circumcentre always lies either inside the triangle or on its boundary. In the case of an obtuse triangle, the circumcentre lies outside the triangle, but it still lies on the extension of one of the triangle’s sides.
### 2. How can the circumcentre be calculated?
The circumcentre can be calculated using various methods, including:
• Using the intersection of perpendicular bisectors: Find the equations of the perpendicular bisectors of two sides and solve them simultaneously to find the point of intersection, which is the circumcentre.
• Using the circumradius formula: If the coordinates of the triangle’s vertices are known, the circumcentre can be calculated using the circumradius formula, which involves finding the intersection of the perpendicular bisectors and calculating the distance between the circumcentre and any vertex.
### 3. Can a triangle have multiple circumcentres?
No, a non-degenerate triangle can have only one circumcentre. The circumcentre is a unique point that satisfies the conditions of being equidistant from the vertices and the intersection of the perpendicular bisectors.
### 4. What is the relationship between the circumcentre and incenter of a triangle?
The incenter of a triangle is the point where the angle bisectors of the triangle’s interior angles intersect. Unlike the circumcentre, the incenter does not necessarily lie inside the triangle. The circumcentre and incenter are distinct points with different properties and applications.
### 5. Can the circumcentre coincide with one of the triangle’s vertices?
Yes, in the case of an equilateral triangle, the circumcentre coincides with all three vertices. This is because an equilateral triangle has all sides equal in length, and the perpendicular bisectors of its sides intersect at the same point, which is equidistant from all three vertices.
## Summary
The circumcentre of a triangle is a fascinating concept that holds significant properties and applications. It is the point where the perpendicular bisectors of the triangle’s sides intersect and is equidistant from the triangle’s vertices. The circumcentre has unique existence, plays a role in triangle construction, triangulation algorithms, optimal location determination, and geometric analysis. Understanding the properties and applications of the circumcentre enhances our knowledge of triangles and their behavior in various contexts. |
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# Algebraic Identities of Polynomials
Algebraic Identities are defined for the algebraic expressions. Algebraic expressions contain variables(a, b, c, x, y, z, etc), numbers(0, 1, 2, 3, 4 …etc) and operators(+, -, *, /….etc). An Algebraic Expression may contain only constants (1, 2, 3, 4, etc), or only variables (x, y, z, etc), or both constant and variable together (5xy, 4p3). Algebraic Identities are basically those Mathematical Equations that make calculations easy in real life.
For example: Consider multiplying two numbers like “989” and “1011”. Now, this is a long calculation, but if you know some identities which suit this kind of problem, It can be solved easily. Before going into detail about algebraic identities, first, let’s see what is an Identity:
### What is an Identity?
Identity is a relation between two or more than two mathematical expressions, such that they produce the same value for all values of variables. In simple words, it can also be said that the L.H.S of any equation becoming identically equal to the R.H.S, for all values of variables explains an Identity.
Let’s look at this expression given below,
(x + 2)(x + 4) = x2 + 6x + 8
Evaluate both sides RHS and LHS of this equation for different values of x,
1. x = 5
LHS: (x + 2)(x + 4) = (5 + 2)(5 + 4) = 63
RHS: x2 + 6x + 8 = 52 + 6(5) + 8 = 25 + 30 + 8 = 63
Thus, both sides of this expression are equal for x = 5.
2. x = 10
LHS: (x + 2) (x + 4) = (10 + 2) (10 + 4) = (12)(14) = 168
RHS: x2 + 6x + 8 = 102 + 6(10) + 8 = 100 + 60 + 8 = 168
Thus, both sides of this expression are equal for x = 10.
If we keep on trying this out with different values of x, we will see that L.H.S and R.H.S are equal for every value of x. Such an expression that is true for every value of variables present in it is called Identity.
Note: An equation is only true for some values of variables present in it
For example:
a2 + 3a + 2 = 132
a = 10 satisfies this identity, but a = 5 or – 7 cannot.
## Types of Algebraic Expressions
Expressions can be of different types depending on how many terms they contain. There are four different types of expressions that make up identities.
### Monomial Expressions
An expression that contain only one term is called as a Monomial Expression.
For example: 16z2, 8xy, -7m, 11…. etc.
Note: A Monomial Expression can be only a constant, a variable or a combination of both constants and Variables.
For Example: 4, x3, 15x2
### Binomial Expressions
An expression containing only two terms is called a Binomial Expression.
For example: x + y, 2x + 5z, x2 + 10 .. etc.
### Proof of Binomial identities:
Identity 1: (a + b)2 = a2 + 2ab + b2
Proof: L.H.S. = (a + b)2
L.H.S. = (a + b) (a + b)
By multiplying each term, we get,
L.H.S = a2 + ab + ab + b2
L.H.S. = a2 + 2ab + b2
L.H.S. = R.H.S.
Identity 2: (a – b)2 = a2 – 2ab + b2
Proof: By taking L.H.S.,
(a – b)2 = (a – b) (a – b)
(a – b)2 = a2 – ab – ab + b2
(a – b)2 = a2 – 2ab + b2
L.H.S. = R.H.S.
Hence, proved.
Identity 3: a2 – b2 = (a + b) (a – b)
Proof: By taking R.H.S and multiplying each term.
(a + b) (a – b) = a2 – ab + ab – b2
(a + b) (a – b) = a2 – b2
Or
a2 – b2 = (a + b) (a – b)
L.H.S. = R.H.S.
Hence proved.
### Trinomial Expression
An expression containing only three terms is called as Trinomial Expression.
For example: 2a + 3b – 5, a2b – ab2 + b2
Identity: (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ac
Proof: Taking L.H.S.
(a+b+c)2= (a+b+c) × (a+b+c)
Using Distributive Property:
(a+b+c)2= a (a+b+c) +b (a+b+c) +c (a+b+c)
= a2+ab+ac+ab+b2+bc+ca+cb+c2
Rearranging the following:
(a+b+c)2= a2+b2+c2+2ab+2bc+2ca
Hence, L.H.S. = R.H.S.
### Polynomials
It is a generalization of all three and other types of expression. An expression containing, one or more terms with a non-zero coefficient (with variables having non-negative exponents) is called a polynomial. A polynomial may contain any number of terms, one or more than one.
Ex: x + y, 2a + 3b – 5, 16z2, 2a + 3b – 5 + z.
Now we’re ready for looking into Algebraic Identities.
## Algebraic Identities
It is very important to learn about the basic algebraic identities which are also known as the standard identities.
Some other Identities:
Let’s look at some examples which use these identities.
### Sample Problems
Question 1: Find out using identities mentioned above, (4x + 3y)2
Solution:
This can be found out using the identity of (a + b)2 = a2 + b2 + 2ab.
(4x + 3y)2 = (4x)2 + (3y)2 + 2(4x)(3y)
= 16x2 + 9y2 + 24xy
Question 2: Find the values 992
Solution:
Multiplying 99 with 99 will take time and calculation. We can formulate this problem in a form that is easier to calculate.
We have seen the identity, (a – b)2 = a2 + b2 – 2ab.
So, 992 = (100 – 1)2 = 1002 + 12 – 2(100)(1)
= 10000 + 1 -200
= 9801
Question 3: Find out 9832 – 172
Solution:
This can take a lot of calculation if we do it the traditional way. We should use the identities to solve this.
We can use a2 – b2 = (a + b)(a -b)
So, 9832 – 172 = (983 + 17)(983 – 17)
= (1000)(966)
= 966000
Question 4: Find out
Solution:
We can use a2 – b2 = (a + b)(a -b)
Question 5: Find out 10112
Solution:
This problem can be solved using multiple identities. Let’s solve it using the identity with three variables.
(a + b + c ) 2 = a2 + b2 + c2 + 2ab + 2bc + 2ca
10112 = (1000 + 10 + 1)2 = 10002 + 102 + 12 + 2(1000)(10) + 2(10)(1) + 2(1000)
= 1000000 + 100 + 1 + 20000 + 20 + 2000
= 1022121
Related Tutorials |
# Mathematical Progressions
May 06, 2019 2 minutes
A mathematical progression is a sequence of numbers such that all the members of the sequence are governed by certain rule, which defines the relation between consecutive terms. There are three types of mathematical progressions:
#### 1. Arithmetic Progression
An arithmetic progression (AP) or arithmetic sequence is a sequence of numbers such that the difference between the consecutive terms is constant.
For an AP with common difference as d and first term as $a_1$, the terms of the progression can be represented as: $a_1, a_1+d, a_1+2d, a_1+3d … a_1+(n-1)d$
Consider the following sequence for example:
• {1,2,3,4,5}: AP with common difference as 1
• {2,4,6,8,10}: AP with common difference as 2
• {5,10,15,20,25}: AP with common difference as 5
For a sequence of terms ${a_1, a_2, … . a_n}$ with common difference as d
• The $n^{th}$ term is given by : $a_n = a_1 + (n-1)d$
• Sum of first n terms of the sequence $S_n = \frac{n}{2}(a_1+a_n)$
• The sum of first n terms can also be represented as $S_n = \frac{n}{2}\bigl(2a_1 + (n-1)\bigr)d$
• $N^{th}$ term can be represented as $Sum(n) - Sum(n-1)$
• The middle term of three consecutive terms of an AP is the mean of the first and last. $a_2 = \frac{a_1 + a_3}{2}$
#### 2. Geometric Progression
A geometric progression (GP) or geometric sequence is a sequence of numbers such that the ratio between the consecutive terms is constant.
On similar lines as mentioned above for AP, consider the first term to be a and common ratio to be r, the terms of GP can be represented as: $a, ar, ar^2, ar^3 … ar^{n-1}$
Consider the following sequence for example:
• {2,4,8,16,32}: GP with first term as 2 and common ratio as 2
• {5,10,20,40,80}: GP with first term as 5 and common ratio as 2
Some useful tips on GP:
• The $n^{th}$ term is given by : $a_n = ar^{n-1}$
• Sum of first n terms $S_n = \frac{ar^n - 1}{r-1}; r \ne 1$
#### 3. Harmonics Progression
Harmonic Progression or harmonic series is a sequence of numbers that originally form an arithmetic progression. Consider the terms $a, a+d, a+2d … a+(n-1)d$ of an arithmetic progression with first term as a and common difference as d, the terms of the HP will be given by: $$\frac{1}{a}, \frac{1}{a+d}, \frac{1}{a+2d}… \frac{1}{a+(n-1)d}$$
Consider the following examples:
• $1,\frac{1}{2}, \frac{1}{3},\frac{1}{4},\frac{1}{5}$; {1,2,3,4,5} forms an AP with a=1 and d=1
• $\frac{1}{2},\frac{1}{4},\frac{1}{6},\frac{1}{8},\frac{1}{10}$; {2,4,6,8,10} forms an AP with a=2 and d=2
• $\frac{1}{5},\frac{1}{10},\frac{1}{15}, \frac{1}{20},\frac{1}{25}$; {5,10,15,20,25} forms an AP with a=5 and d=10 |
# Difference between revisions of "2019 AMC 10B Problems/Problem 9"
## Problem
The function $f$ is defined by $$f(x) = \lfloor|x|\rfloor - |\lfloor x \rfloor|$$for all real numbers $x$, where $\lfloor r \rfloor$ denotes the greatest integer less than or equal to the real number $r$. What is the range of $f$?
$\textbf{(A) } \{-1, 0\} \qquad\textbf{(B) } \text{The set of nonpositive integers} \qquad\textbf{(C) } \{-1, 0, 1\} \qquad\textbf{(D) } \{0\} \qquad\textbf{(E) } \text{The set of nonnegative integers}$
## Solution
There are 4 cases we need to test here:
Case 1: x is a positive integer. WLOG, assume x=1. Then f(1) = 1 - 1 = $0$.
Case 2: x is a positive fraction. WLOG, assume x=0.5. Then f(0.5) = 0 - 0 = $0$.
Case 3: x is a negative integer. WLOG, assume x=-1. Then f(-1) = 1 - 1 = $0$.
Case 4: x is a negative fraction. WLOG, assume x=-0.5. Then f(-0.5) = 0 - 1 = $-1$.
Thus the range of function f is $\boxed{\textbf{(A) } \{-1, 0\}}$
## Solution 2
It is easily verified that when $x$ is an integer, then $f(x)$ is zero. We need only to consider the case when $x$ is not.
When $x$ is a positive number, $\lfloor x\rfloor \geq 0$, so $$f(x)=\lfloor|x|\rfloor-|\lfloor x\rfloor|$$ $$=\lfloor x\rfloor-\lfloor x\rfloor$$ $$=\textbf{0}$$
When $x$ is a negative number, let $x=-a-b$ be composed of integer part $a$ and decimal part $b$ (both $\geq 0$): $$f(x)=\lfloor|-a-b|\rfloor-|\lfloor -a-b\rfloor|$$ $$=\lfloor a+b\rfloor-|-a-1|$$ $$=a-(a+1)=\textbf{-1}$$
Thus, the range of f is $\boxed{\textbf{(A) } \{-1, 0\}}$
>>> Intelligence_Inc
Note: One could solve the case of $x$ as a negative non-integer this way: $$f(x)=\lfloor|x|\rfloor-|\lfloor x\rfloor|$$ $$=\lfloor -x\rfloor-|-\lfloor -x\rfloor-1|$$ $$=\lfloor -x\rfloor-(\lfloor -x\rfloor+1) = \textbf{-1}$$ |
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Problem 11
# Aragorn the Great has boasted to his hordes of followers that many a notorious villain has fallen to his awesome sword: His total of 560 victims consists of evil sorcerers, trolls, and orcs. These he has slain with a total of 620 mighty thrusts of his sword; evil sorcerers and trolls each requiring two thrusts (to the chest) and orcs each requiring one thrust (to the neck). When asked about the number of trolls he has slain, he replies, "I, the mighty Aragorn, despise trolls five times as much as I despise evil sorcerers. Accordingly, five times as many trolls as evil sorcerers have fallen to $$\mathrm{m} \mathrm{y}$$ sword!" How many of each type of villain has he slain?
Expert verified
Aragorn has slain 10 evil sorcerers, 50 trolls, and 500 orcs.
See the step by step solution
## Step 1: Equation for Total Victims
We are given that Aragorn has slain 560 victims in total, which include evil sorcerers, trolls, and orcs. Let x be the number of evil sorcerers, y be the number of trolls, and z be the number of orcs. Then we have the following equation for the total victims: x + y + z = 560
## Step 2: Equation for Total Sword Thrusts
We are given that Aragorn used a total of 620 sword thrusts to slay the evil sorcerers, trolls, and orcs. Evil sorcerers and trolls require two sword thrusts each, while orcs only require one. Thus, we can write another equation representing the total sword thrusts: 2x + 2y + z = 620
## Step 3: Relationship between Evil Sorcerers and Trolls
Finally, we are given that Aragorn has slain five times as many trolls as evil sorcerers. This can be written as the following equation: y = 5x
## Step 4: System of Equations
We now have a system of three linear equations with three unknowns: 1. x + y + z = 560 2. 2x + 2y + z = 620 3. y = 5x
## Step 5: Solving the System of Equations
To solve this system of equations, we can use the substitution method. We can substitute the third equation into the first and second equations: 1. x + 5x + z = 560 2. 2x + 2(5x) + z = 620 Simplifying this gives us: 1. 6x + z = 560 2. 12x + z = 620 Now solve the first equation for z: z = 560 - 6x Now substitute this into the second equation: 12x + (560 - 6x) = 620 Simplify and solve for x: 6x = 60 x = 10 Now we can find the values for y and z using the values we found for x: y = 5x = 5(10) = 50 z = 560 - 6x = 560 - 60 = 500
## Step 6: Final Solution
So, Aragorn has slain 10 evil sorcerers, 50 trolls, and 500 orcs.
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## Precalculus (6th Edition) Blitzer
The required solution is $4{{x}^{3}}+6{{x}^{2}}h+4x{{h}^{2}}+{{h}^{3}}$.
Let us consider the provided ratio: $\frac{f\left( x+h \right)-f\left( x \right)}{h}$ Now, put the provided function in the above ratio: \begin{align} & \frac{\left( {{\left( x+h \right)}^{4}}+7 \right)-\left( {{x}^{4}}+7 \right)}{h}=\frac{{{\left( x+h \right)}^{4}}+7-{{x}^{4}}-7}{h} \\ & =\frac{{{\left( x+h \right)}^{4}}-{{x}^{4}}}{h} \\ & =\frac{{{x}^{4}}+4{{x}^{3}}h+6{{x}^{2}}{{h}^{2}}+4x{{h}^{3}}+{{h}^{4}}-{{x}^{4}}}{h} \\ & =\frac{h\left( 4{{x}^{3}}+6{{x}^{2}}h+4x{{h}^{2}}+{{h}^{3}} \right)}{h} \end{align} So, $\frac{\left( {{\left( x+h \right)}^{4}}+7 \right)-\left( {{x}^{4}}+7 \right)}{h}=4{{x}^{3}}+6{{x}^{2}}h+4x{{h}^{2}}+{{h}^{3}}$ Thus, the ratio is $4{{x}^{3}}+6{{x}^{2}}h+4x{{h}^{2}}+{{h}^{3}}$. It is calculated by putting the function into the given ratio and the value obtained matches with the binomial expansion formula. |
## Engage NY Eureka Math Kindergarten Module 5 Lesson 23 Answer Key
### Eureka Math Kindergarten Module 5 Lesson 23 Problem Set Answer Key
Question 1.
Robin sees 5 apples in a bag and 10 apples in a bowl. Draw a picture to show how many apples there are.
Explanation:
I drew 10 aplles in a bowl and 5 apples in a bag.
Question 2.
Write a number bond and an addition sentence to match your picture.
___ ___ ___
Explanation:
The above number bond tells us about my picture of apples.
Question 3.
Sam has 13 toy trucks. Draw and show the trucks as 10 ones and some ones.
Explanation:
I drew 10 toy cars and 3 toy cars.
Question 4.
Write a number bond and an addition sentence to match your picture.
___ ___ ___
Explanation:
The above number bond tells us about picture of Sam’s toy trucks.
Question 5.
Our class has 16 bags of popcorn. Draw and show the popcorn bags as 10 ones and some ones.
Explanation:
I drew 10 popcorn bags and 6 popcorn bags.
Question 6.
Write a number bond and an addition sentence to match your picture.
___ ___ ___
Explanation:
The above number bond tells us about my picture of popcorn bags.
### Eureka Math Kindergarten Module 5 Lesson 23 Exit Ticket Answer Key
Question 1.
There are 12 balls. Draw and show the balls as 10 ones and some ones.
Explanation:
I drew 10 balls and 2 balls.
Question 2.
Write a number bond to match your picture.
Explanation:
The above number bond tells us about my picture of balls.
Question 3.
_______ _______ _______
12 = 10 + 2
Explanation:
The addition sentence that match the picture and the number bond is 12 = 10 + 2.
### Eureka Math Kindergarten Module 5 Lesson 23 Homework Answer Key
Question 1.
Bob bought 7 sprinkle donuts and 10 chocolate donuts. Draw and show all of Bob’s donuts.
Explanation:
I drew 10 chocolate donuts and 7 sprinkle donuts.
Question 2.
_______ _______ _______
17 = 10 + 7
Explanation:
The addition sentence that matches with the number bond and the picture is 17 = 10 + 7.
Question 3.
Fill in the number bond to match your sentence.
Explanation:
The above number bond tells us about picture of bob’s donuts.
Question 4.
Fran has 17 baseball cards. Show Fran’s baseball cards as 10 ones and some ones. |
# Thread: A geometric problem I am having issues with...
1. ## A geometric problem I am having issues with...
This is from an online worksheet. I cannot work this problem out.... Use the following diagram to answer the next two questions... Correct to the nearest tenth, the value of x is A. 5.2 B. 5.3 C. 5.4 D. 5.5 .................................................. .................................................. .................................................. ..... 2. Correct to the nearest half of a degree, the measure of angle ADB is: A. 30.5° B. 32.0° C. 34.5° D. 36.0° Thank you for the help!
2. Originally Posted by M23
This is from an online worksheet. I cannot work this problem out.... Use the following diagram to answer the next two questions... Correct to the nearest tenth, the value of x is A. 5.2 B. 5.3 C. 5.4 D. 5.5
Use Pythagorean theorem in RIGHT TRIANGLE PBC,
$\displaystyle PC^2=PB^2+BC^2$
$\displaystyle PC^2=9^2+5^2$
PC=10.3
x = 10.3 - 5 = 5.3
Spoiler:
Pythagorean Theorem in Right Triangle
$\displaystyle a^2 = b^2+c^2$
$\displaystyle \Rightarrow(Hypotenuse)^2 = (first\;side)^2+(second \;side)^2$
3. Originally Posted by M23
This is from an online worksheet. I cannot work this problem out.... Use the following diagram to answer the next two questions... 2. Correct to the nearest half of a degree, the measure of angle ADB is: A. 30.5° B. 32.0° C. 34.5° D. 36.0° Thank you for the help!
IN Right triangle PBC,
$\displaystyle \tan P = \frac{5}{9}$
$\displaystyle P = 29.05^\circ$
$\displaystyle \angle C = 180 - 90 -29.05 = 60.95^\circ$
$\displaystyle \angle D = \frac{ 60.95}{2}=30.47=30.5^\circ$
Spoiler:
$\displaystyle \angle D = \frac{\angle C}{2}$
4. Thank you soooo much. |
This math lesson is designed for 6-12 years old children to help them learn the cube numbers and notation of cubes using Montessori bead bars.
In our previous video lesson, we learned about how to find squares of a number using Montessori bead bars. In this video, we will learn how to multiply and make cube notations using Montessori bead bars.
## What is the Notation of a Cube?
When a number is raised to the power of three, it is called the cube of that number. The notation used for a cube of a number is to write the number with a small 3 exponent to its upper right.
For example, if we want to find the cube of the number 2, we write it as 2³. This notation indicates that we are taking 2 and multiplying it by itself three times. The result is 8, which is the cube of 2.
Similarly, if we want to find the cube of any number ‘a’, we write it as ‘a³’. This notation helps to quickly identify the operation that needs to be performed to find the cube of a number.
## Why are Cube Numbers called Cubed Numbers?
Cube numbers (or cubed numbers) are so called because they are used to calculate the volume of a cube. A cube is a three-dimensional shape with sides of equal length, width, and height. To calculate its volume, we multiply the side length by itself twice, or “cube” it.
For example, a cube with a side length of 2 cm has a volume of 8 cm³ (since 2³ = 8). If we knew a cube had a volume of 27 cm³, we could deduce that each side measures 3 cm (since 3³ = 27).
This connection between cube numbers and the volume of a cube is why they are called cube numbers.
### How Notation of Cubes Are Introduced in a Montessori Classroom?
In a Montessori classroom, the notation of cubes is introduced using the Montessori bead bar. As shown in the video, children can explore cubes by creating them with bars. Bead bars of the same length can be used to make various-sized cubes.
## How to Find a Cube of the Number or Square Notation in a Montessori Way?
This notation method is a simple yet effective way for children to understand and remember the concept of squares. To find the square of a number, follow the below steps:
1. Invite the child to the table along with the Montessori bead bars.
2. Tell them today we will find the cube of a number using Montessori bead bars.
3. As shown in the video, let’s take 3 bead bars string 3 times and arrange them together.
4. Now ask the child to count the bead bars both vertically and horizontally.
5. Note the numbers and present them as 3x3x3 in the notebook.
6. Now, to notate a cube, children can count the number of beads used to create the cube, either counting all the beads or skip counting. For example, if a child creates a square using five bead bars of the same length, they can notate it as 33. The superscript of 3 indicates that the shape is a cube.
8. Encourage the child to try finding the cube of a number and write its notation with different numbered bars.
## Benefits of Finding Square of a Number or Notation of Square
Here are five benefits of finding the cubes of a number:
Here are five benefits of finding the cubes of a number:
1. Simplifies algebraic equations: Notating cubes makes it easier to write and understand equations involving cube variables.
2. Helps in geometry: Cube numbers can be used to calculate the volume of a cube, which is a three-dimensional shape with sides of equal length, width, and height.
3. Fosters quick mental math: Knowing the cubes of small numbers can aid in quick calculations and mental math.
4. Useful in scientific calculations: Cubes can be used to determine distances between objects in space, among other things.
5. Aids in understanding exponents: Knowing cubes will help you understand squares and higher powers of exponents.
## List of Square Numbers
Here are the squares of numbers 1-20:
• 1x1x1 = 1³ = 1
• 2x2x2 = 2³ = 8
• 3x3x3 = 3³ = 27
• 4x4x4 = 4³ = 64
• 5x5x5 = 5³ = 125
• 6x6x6 = 6³ = 216
• 7x7x7 = 7³ = 343
• 8x8x8 = 8³ = 512
• 9x9x9 = 9³ = 729
• 10x10x10 = 10³ = 1000
• 11x11x11 = 11³ = 1331
• 12x12x12 = 12³ = 1728
• 13x13x13 = 13³ = 2197
• 14x14x14 = 14³ = 2744
• 15x15x15 = 15³ = 3375
• 16x16x16 = 16³ = 4096
• 17x17x17 = 17³ = 4913
• 18x18x18 = 18³ = 5832
• 19x19x19 = 19³ = 6859
• 20x20x20 = 20³ = 8000
In algebra, cubing is an important operation that is used in many equations and formulas. By using the notation of cubes, it becomes easier to write and understand these equations.
Ask the child to write cube notations as shown in the video, and help them get a better understanding of the concept with this fun and interactive method.
## FAQs
• How do you write a cube in math?
The notation used for finding a cube of a number is to write the number with a small 3 exponent to its upper right. For example, the cube of 2 is written as 2³.
• What are the benefits of finding the cubes of a number?
There are several benefits of finding the cubes of a number. It facilitates quick mental math, simplifies algebraic equations, and aids in geometry. In addition, cubes can be used to understand other exponents, such as squares and higher powers.
• What is the formula of a cube?
The formula for finding the volume of a cube is given by: Volume of Cube = Side³. Where ‘Side’ is the length of one side of the cube. To use the formula, simply cube the length of any one side of the cube to find its volume.
Tags
• elementary level
• Math |
# Solving Quadratic Equations by Factorisation
The process of writing an expression as a product of two or more common factors is called method of factorization. e.g.
1. ${x^2} + 5x + 6 = \left( {x + 2} \right)\left( {x + 3} \right)$
2. $5{x^2} + 8x = x\left( {5x + 8} \right)$
3. $30 = 2 \times 3 \times 5$
In the above examples, $\left( {x + 2} \right)\left( {x + 3} \right)$are the factors of expression ${x^2} + 5x + 6$, $x\left( {5x + 8} \right)$are the factors of $5{x^2} + 8x$ and $2 \times 3 \times 5$are the factors of $30$.
While solving the quadratic equation by the method of factorization, we have the following steps:
• Convert the quadratic equation in standard form, if necessary i.e. $a{x^2} + bx + c = 0$, where $a \ne 0$
• Multiply coefficient of ${x^2}$ with constant terms, we get $a \times c = ac$.
• Now try to find two numbers whose products is $ac$ and sum or difference is equal to $b$ (coefficient of $x$).
• Factorise the given expression on L.H.S.
• Equate each factor equal to zero.
• We get the required roots, say${x_1}$, ${x_2}$.
Example:
Solve the equation by factorization method.
Solution:
The given equation in standard form is $5{x^2} - 11x + 6 = 0$
Multiply coefficient of${x^2}$and the constant term, we get $5 \times 6 = 30$
Divide $30$ into two parts such that their difference or sum is $11$
Possible factors of $30$ Sum or Difference of factors $30 \times 1 = 30$ $30 - 1 = 29,{\text{ }}30 + 1 = 31$ (not possible) $15 \times 2 = 30$ $15 - 2 = 13,{\text{ 15}} + 2 = 17$ (not possible) $10 \times 3 = 30$ $10 - 3 = 7,{\text{ 1}}0 + 3 = 13$ (not possible) $6 \times 5 = 30$ $6 - 5 = 1,{\text{ 6}} + 5 = 11$ (possible)
Therefore, $5{x^2} - 11x + 6 = 0$
$5{x^2} - \left( {5 + 6} \right)x + 6 = 0$
$5{x^2} - 5x - 6x + 6 = 0$
$5x\left( {x - 1} \right) - 6\left( {x - 1} \right) = 0$
$\left( {x - 1} \right)\left( {5x - 6} \right) = 0$
Either $x - 1 = 0$ or $5x - 6 = 0$
$x = 1$ or $x = \frac{6}{5}$
Example:
Solve the equation by factorization method.
Solution:
The given equation in standard form is $4{x^2} + 8x = 0$
Here, the constant term is absent; its factorization is very simple.
Taking common $4x$, we get
$4x\left( {x + 2} \right) = 0$
Either $4x = 0$ or $x + 2 = 0$
$x = 0$ or $x = - 2$ |
# Bihar Board Class 9th Maths Solutions Chapter 7 Triangles Ex 7.1
Bihar Board Class 9th Maths Solutions Chapter 7 Triangles Ex 7.1 Textbook Questions and Answers.
## BSEB Bihar Board Class 9th Maths Solutions Chapter 7 Triangles Ex 7.1
Question 1.
In quadrilateral ACBD, AC = AD and AB bisects ∠A (see figure). Show that ∆ ABC ≅ ∆ ABD. What can you say about BC and BD?
Solution:
Now, in As ABC and ABD, we have
∠CAB = ∠BAD (∵ AB bisects ∠A]
and, AB = AB [Common]
∴ By SAS congruence criterion, we have
∆ ABC ≅ ∆ ABD
⇒ BC = BD
[∵ Corresponding parts of congruent triangles are equal]
Question 2.
ABCD is a quadrilateral in which AD = BC and ∠DAB = ∠CBA (see figure). Prove that
(i) ∆ ABD ≅ ∆BAC
(ii) BD = AC
(iii) ∠ABD = ∠BAC.
Solution:
In ∆s ABD and BAC, we have
∠DAB = ∠CBA lGiven]
AB = AB [Common]
By SAS criterion of congruence, we have A ABD = A BAC, which proves (i)
⇒ BD = AC
and, ∠ABD = ∠BAC, which proves (ii) and (iii)
[∵ Corresponding parts of congruent triangles are equal]
Question 3.
AD and BC are equal „ perpendiculars to a line segment AB (see figure). Show that CD bisects AB.
Solution:
Since AB and CD intersect at O. Therefore,
∠AOD =∠BOC … (1) [Vertically opp. angles]
In ∆s AOD and BOC, we have
∠AOD = ∠BOC [From(1)]
∠D AO = ∠OBG [Each = 90°]
∴ By AAS congruence criterion, we have ∆ AOD ≅ ∆ BOC
OA = OB
[∵ Corresponding parts of congruent triangles are equal] i.e., O is the mid-point AB.
Hence, CD bisects AB.
Question 4.
1 and m are two parallel lines intersected by another pair of parallel lines p and q (see figure). Show that ∆ ABC ≅ ∆ CDA.
Solution:
Since l and m are two parallel lines intersected by another pair of parallel lines p and q. Therefore, AD || BC and AB || CD ⇒ ABCD is a parallelogram.
i.e., AB = CD
Now, in ∆s ABC and CDA, we have
AB = CD [Prove above]
and AC = AC [Common]
∴ By SSS criterion of congruence ∆ ABC ≅ ∆ CDA.
Question 5.
Line l is the bisector of an angle A and B is any point on l. BP and BQ are perpendiculars from B to the arms of ∠A (see figure). Show that :
(i) ∆ APB ≅ ∆ AQB
(ii) BP = BQ or B is equidistant from the arms of ∠A.
Solution:
In ∆s APB and AQB, we have
∠APB = ∠AQB [∵ Each = 90°]
∠PAB = ∠QAB [∵ AB bisects ∠PAQ]
AB = AB [Common]
By AAS congruence criterion, we have
∆ APB ≅ ∆ AQB, which proves (i)
⇒ BP = PQ
[∵Corresponding parts of congruent triangles are equal]
i.e., B is equidistant from the arms of ∠A, which proves
Question 6.
In figure, AC = AE, AB = AD and ∠BAD = ∠EAC. Show that BC = DE.
Solution:
∠BAC = ∠DAE
[∵ ∠BAD = ∠EAC ⇒ ∠BAD + ∠DAC = ∠EAC + ∠DAC ⇒ ∠BAC =∠DAE]
and, AC = AE [Given]
∴ By SAS criterion of congruence, we have
⇒ BC = DE
[∵ Corresponding parts of congruent triangles are equal]
Question 7.
AB is a line segment and P is its mid-point. D and E are points on the same side of AB such that ∠BAD = ∠ABE and ∠EPA = ∠DPB (see figure). Show that
(i) ∆ DAP ≅ ∆ EBP
Solution:
We have, ∠EPA = ∠DPB
⇒ ∠EPA + ∠DPE = ∠DPB + ∠DPE’
⇒ ∠DPA = ∠EPB … (1)
Now, in ∆s EBP and DAP, we have
∠EPB = ∠DPA [From (1)]
BP = AP [Given]
and, ∠EBP = ∠DAP [Given]
So, by ASA criterion of congruence, we have
∆ EBP ≅ ∆ DAP
[∵ Corresponding parts of congruent triangles are equal]
Question 8.
In right triangle ABC, right angled at C, M is the mid-point of hypotenuse AB. C is joined to M and produced to a point D such that DM = CM. Point D is joined to point B (see figure). Show that:
(i) ∆ AMC ≅ ∆ BMD
(ii) ∠DBC is a right angle
(iii) ∆ DBC ≅ ∆ ACB
(iv) CM = $$\frac { 1 }{ 2 }$$ AB.
Solution:
(i) In ∆s AMC and BMD, we have
AM = BM [∵ M is the mid-point of AB]
∠AMC = ∠BMD [Vertically opp. ∠s]
and, CM = MD [Given]
∴ By SAS criterion of congruence, we have
∴ ∆ AMC ≅ ∆ BMD
(ii) Now, ∆ AMC ≅ ∆ BMD
⇒ BD = CA and ∠BDM = ∠ACM … (1)
[∵Corresponding parts of congruent triangles are equal]
Thus, transversal CD cuts CA and BD at C and D respectively such’that the alternate angles ∠BDM and ∠ACM are equal. Therefore, BD || CA.
⇒ ∠CBD + ∠BCA = 180°
[∵ Sum of the interior angles on the same side of ” transversal = 180°]
⇒ ∠CBD + 90° = 180° [∵ ∠BCA = 90°]
⇒ ∠DBC = 90°.
(iii) Now, in ∆s DBC and ACB, we have
BD = CA [From (1)]
∠DBC = ∠ACB [∵ Each = 90°)
BC = BC [Common]
By SAS criterion of congruence, we have ∆ DBC ≅ ∆ ACB.
(iv) CD = AB
[∵ Corresponding parts of congruent triangles are equal]
⇒ $$\frac { 1 }{ 2 }$$CD = $$\frac { 1 }{ 2 }$$AB ⇒ CM = $$\frac { 1 }{ 2 }$$AB. |
# What are the coordinates of the vertex of the parabola whose equation is y = 3(x - 2)^2 + 5?
Apr 12, 2015
The answer is: $V \left(2 , 5\right)$.
There are two ways.
First:
we can remember the equation of the parabola, given the vertex $V \left({x}_{v} , {y}_{v}\right)$ and the amplitude $a$:
$y - {y}_{v} = a {\left(x - {x}_{v}\right)}^{2}$.
So:
$y - 5 = 3 {\left(x - 2\right)}^{2}$ has vertex: $V \left(2 , 5\right)$.
Second:
we can make the counts:
$y = 3 \left({x}^{2} - 4 x + 4\right) + 5 \Rightarrow y = 3 {x}^{2} - 12 x + 17$
and, remembering that $V \left(- \frac{b}{2 a} , - \frac{\Delta}{4 a}\right)$,
$V \left(- \frac{- 12}{2 \cdot 3} , - \frac{{12}^{2} - 4 \cdot 3 \cdot 17}{4 \cdot 3}\right) \Rightarrow V \left(2 , 5\right)$.
Apr 12, 2015
Vertex is $\left(2 , 5\right)$
Method
Use the form: ${\left(x - h\right)}^{2} = 4 a \left(y - k\right)$
This parabola has vertex at $\left(h , k\right)$
And its major axis is along the $y - \text{axis}$
In our case we have, $y = 3 {\left(x - 2\right)}^{2} + 5$
$\implies 3 {\left(x - 2\right)}^{2} = y - 5$
$\implies {\left(x - 2\right)}^{2} = \frac{1}{3} \left(y - 5\right)$
So, the vertex is $\left(2 , 5\right)$
Worthy of note
When the equation is of the form: ${\left(y - k\right)}^{2} = 4 a \left(x - h\right)$
The vertex is at $\left(h , k\right)$ and the parabola lies along the $x - \text{axis}$ |
CONNECT: Areas, Perimeters
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1 CONNECT: Areas, Perimeters 1. AREAS OF PLANE SHAPES A plane figure or shape is a two-dimensional, flat shape. Here are 3 plane shapes: All of them have two dimensions that we usually call length and width (or sometimes height). Plane shapes do not have thickness, which means we can draw them on paper. The amount of space inside each shape or the amount of space each figure occupies is called the area of that shape. Units to measure area When you hear the word area, you may think of school maths when you had to find the area of different shapes, mostly rectangles and squares. But we can actually find the area of ANY shape, even the strange one above. To do this we firstly need to think about the basic unit of area. Try to think of a small shape that, if it was repeated over and over again, perhaps turned upside down or back-to-front, would cover each of the shapes above. Using a shape like this is called tessellating and examples can be found in patchwork, tiling, mosaics and mosques. Designs range from simple to ornate and can be very attractive. I have copied some examples over the page. 1
2 Retrieved January 22, 2013 from and Retrieved January 22, 2013 from and But what if we need to compare the areas of our shapes on the first page? Say we want to cover them all perhaps or, even worse these are the shapes of your late great aunt s farm paddocks and she has left them to the family who is going to get the biggest share?! The small shape that has been accepted throughout the world for tessellating to determine areas is the square. In the Metric System of measurement, the unit of length is the metre and so for area, the unit is the square metre written as m 2. Smaller measures are given in square centimetres (cm 2 ) or even smaller square millimetres (mm 2 ) and large measures are given in hectares (ha). Note that ha doesn t have the superscript 2 as it is already a square unit 1ha measures an area 100m by 100m, that is, 1ha = m 2. 2
3 100m 100m 1ha To be able to work out the areas of the shapes above, we could draw a grid of squares on each shape and count them. (We would need to approximate for the irregular shapes.) The area of any rectangle With regular shapes, the area is more straightforward. To find the area of this rectangle, we can cover it with a grid of squares with length 1cm and width 1cm, that is they are each square centimetres (1cm 2 ), and count them. In this rectangle, there are 12 square centimetres, so its area is 12 cm 2. Is there a faster way to find the area of this rectangle? Yes! We know that the rectangle is 6cm long and 2cm wide, so we can fit two rows each of 6 centimetre squares in the complete rectangle. So, its area is 12cm 2. We can also calculate this as 6cm x 2cm = 12 cm 2. We can generalise this with a formula. If we let the length of the rectangle be l units and the width of the rectangle be w units, then to calculate the area (A units 2 ) we multiply the length by the width and so we have: A = l x w units 2 3
4 We can apply this formula to any rectangle as long as we know its dimensions (the size of its length and width). You may have seen this formula before, using b (represents breadth) instead of w width and breadth are the same and in this case, the area would be A = l b units 2. Both formulas give the same result. To find the area of the rectangle given above, we know that l = 6cm and w = 2cm, so A = 6cm x 2cm = 12 cm 2. To find the area of any rectangle, the dimensions must be measured using the same unit such as centimetres, kilometres and so on. We cannot find the area of a rectangle where the length is given in kilometres and the width is in metres, for example, we must convert one of those measures to the other. The following diagrams are not drawn to scale. See if you can find the area of each rectangle. You can check your results with the solutions at the end of this resource cm 11mm 1.5cm 4.2cm Areas of other shapes What about triangles? Can you see that the area of one of the triangles formed is half the area of the whole rectangle? So we can say its area is 6cm 2 straight away. 4
5 But what about this one? We could do this: Now we have 2 triangles that are both half their respective rectangles in area. So the area of the complete triangle is 2 cm 2 (for the one on the left) + 4 cm 2 (for the one on the right) which makes 6 cm 2. Here is the formula for the area of a triangle you might remember it from school. With a triangle, we ll refer to the length (of the rectangle) as the base of the triangle and the width (of the rectangle) as the height of the triangle. height base The area of a triangle is given by half the area of a rectangle, so the area is given by A = ½ x b x h units 2, where A units 2 represents the area of the triangle, b units represents the length of the base of the triangle and h units represents the height of the triangle. The height of the triangle must be at right angles to the base of the triangle and the units of measure must be the same. 5
6 Example: Find the area of this triangle. 8.5cm 15cm The area is ½ x 15cm x 8.5cm = 63.75cm 2. Find the area of this triangle: 17mm 6.9mm Notice that in this triangle, the base has to be extended so that we can draw a height at right angles to it. The area can still be calculated in the same way, though. So the area = ½ x 17mm x 6.9mm = 58.65mm 2 Here are some for you to try. You can check your results with the solutions at the end. Calculate the areas of these triangles (the diagrams are not drawn to scale): km 65km 6
7 m 10.1m What about circles? If we know the size of the radius of a circle, we can work out the area. Diagram retrieved 22 January 2013 from The radius is any line joining the centre of the circle to the circumference and is half the diameter. If we let r units be the length of the radius, then the area (given by A units 2 ) is A = πr 2 units 2 π is a Greek letter (pronounced pie in Australia) and is always the same number. It is the answer when you divide the length of the circumference of a circle by the length of the diameter of that circle. It is always and can be calculated correctly to many decimal places. If you use a calculator when you work with π, your answer will always be accurate and you will need to round it as it will contain many decimal places. 7
8 When you want to calculate using A = πr 2, you type π in your calculator (often you need to use the Shift key) followed by x, then the radius followed by x 2, and =. Example: Calculate the area of this circle, correct to 3 decimal places. Area = πr 2 In this circle, the radius is 5cm long, so r = 5cm, and the area is πr 2 = π 5 2 cm 2 = cm cm 2 (rounded to 3 decimal places) Here are two for you to try. Remember, if you are given a diameter instead of a radius, you need to calculate half of it (to work out the radius) before calculating the area. (The diagrams are not to scale.) Calculate the area of each of the following circles:
9 Converting between square units 11mm 4.2cm This rectangle was one of the questions from page 4, where you were required to find its area. In the solutions, I changed the centimetre measurements into mm and calculated the area in mm 2. The area is 462mm 2. Now I m going to calculate the area using cm 2 answers. and we will compare our First, change 11mm to cm, by dividing by 10, so 11mm = cm, = 1.1cm. Area = 4.2cm x 1.1cm = 4.62cm 2 Is this what you expected? The area in mm 2 is 100 times as big as the area in cm 2. This is because we are no longer dealing with one dimension (length) where 1cm = 10mm, but with 2 dimensions (length and width): 10mm 10mm 1cm 2 So, 1cm 2 = 100mm 2 In the same way, 1m 2 = 100cm x 100cm = 10000cm 2 and 1km 2 = 1000m x 1000m = m 2 9
10 Here is an example: I have measured a space for a small window in cm 2, but the builder needs the measurement in mm 2. My measurement was 3 600cm 2. What should I tell the builder? 1cm 2 = 100mm 2, so 3 600cm 2 = x 100mm 2 = mm 2 A second example: I have measured a block of land in m 2 but would like to know how many hectares this is. My measurement is m 2. I have put the solution at the end. If you need help with any of the Maths covered in this resource (or any other Maths topics), you can make an appointment with Learning Development through Reception: phone (02) , or Level 3 (top floor), Building 11, or through your campus. 10
11 Solutions Areas of rectangles, (page 4) cm 1.5cm Area = 3.7cm x 1.5cm = 5.55cm mm 4.2cm To deal with this rectangle, we must make sure both measurements are in the same unit. It is probably easier to make them both mm, so 4.2cm = 4.2 x 10mm, that is 42mm. Now the area is found using the formula. Area = 42mm x 11mm = 462mm 2 (If you calculated the area in cm 2, please see page 9.) 11
12 Areas of triangles (page 6, 7) km 65km Area = ½ x 65km x 8.8km = 286km m 10.1m Area = ½ x 14.1m x 10.1m = m 2 Areas of circles (page 8) 1. Area = π 4 2 cm 2 = cm cm 2 (rounded to 3 decimal places) 12
13 2. Firstly, we are given the diameter instead of the radius, so we must find half of 38cm to get the length of the radius. (38 2 = 19). So the area is Area = π 19 2 cm 2 = cm cm 2 (rounded to 3 decimal places) Converting between square units (page 10) 1ha = m 2 The measurement is m 2. How many lots of m 2 are in m 2? We need to divide by to find the number of ha, and obtain 4.25ha. 13
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# What is the standard form of y=(3-x)(x-1)^2?
Nov 29, 2017
$- {x}^{3} + 5 {x}^{2} - 7 x + 3$
#### Explanation:
$\text{expand the factors and collect like terms}$
${\left(x - 1\right)}^{2} = \left(x - 1\right) \left(x - 1\right) \leftarrow \textcolor{b l u e}{\text{expand using FOIL}}$
$\left(x - 1\right) \left(x - 1\right) = {x}^{2} - 2 x + 1$
$\text{now multiply expansion by factor } \left(3 - x\right)$
$\left(3 - x\right) \left({x}^{2} - 2 x + 1\right)$
$\text{multiply each term in the second factor by each term}$
$\text{in the first factor}$
$\textcolor{red}{3} \left({x}^{2} - 2 x + 1\right) \textcolor{red}{- x} \left({x}^{2} - 2 x + 1\right)$
$= 3 {x}^{2} - 6 x + 3 - {x}^{3} + 2 {x}^{2} - x \leftarrow \textcolor{b l u e}{\text{collect like terms}}$
$= - {x}^{3} + 5 {x}^{2} - 7 x + 3 \leftarrow \textcolor{red}{\text{in standard form}}$
$\text{to express a polynomial in "color(blue)"standard form}$
$\text{start with the term with the largest exponent of the variable}$
$\text{followed by terms of decreasing exponents in descending}$
$\text{order}$ |
# 2006 Romanian NMO Problems/Grade 7/Problem 4
## Problem
Let $A$ be a set of positive integers with at least 2 elements. It is given that for any numbers $a>b$, $a,b \in A$ we have $\frac{ [a,b] }{ a- b } \in A$, where by $[a,b]$ we have denoted the least common multiple of $a$ and $b$. Prove that the set $A$ has exactly two elements.
Marius Gherghu, Slatina
## Solution
We first show that $A$ is finite; for the sake of contradiction, suppose that $A$ is infinite. Let $a \in A$ be the smallest element of $A$. Let $n \in A$ such that $n > a$ and let $g = GCD(n,a)$. Then $\frac{[n,a]}{n-a} = \frac{na}{g(n-a)} = \frac{a}{g} + \frac{a^{2}}{g(n-a)}$ is an integer for arbitrarily large $n$; but $\frac{a^{2}}{g(n-a)} < \frac{a^{2}}{n-a} < 1$ if $n > a^{2}+a$. Therefore $A$ is finite.
Let $x,y \in A$ with $x > y$; let $g = GCD(x,y)$ and $x_{0} = x/g$ and $y_{0} = y/g$. Then $\frac{[x,y]}{x-y} = \frac{xy}{g(x-y)} = \frac{x_{0}y_{0}}{x_{0}-y_{0}}$. Suppose $x_{0} - y_{0} > 1$ and let $p$ be a prime dividing $x_{0} - y_{0}$; $p$ divides $x_{0}$ if and only if $p$ divides $y_{0}$. But $p$ divides neither because $x_{0}$ and $y_{0}$ are relatively prime. Thus $x_{0} - y_{0} = 1$ and $g = x-y$ and $\frac{xy}{(x-y)^{2}} \in A$ for all $x,y \in A$ such that $x > y$.
Let $b$ be the smallest element and let $a$ be the largest element of $A$. Since $\frac{ab}{(a-b)^{2}} \in A$, we have $b \le \frac{ab}{(a-b)^{2}} \le a \implies b \le (a-b)^{2} \le a \implies 0 \le (a-b)^{2}-b \le a-b$ and $a-b$ divides $a$ and $b$ so either $0 = (a-b)^{2}-b$ or $(a-b)^{2}-b = a-b$ and $(a-b)^{2}$ equals $a$ or $b$. Suppose there exists some $n \in A$ such that $b < n < a$. If $(a-b)^{2} = b$, then $\frac{[n,a]}{a-n} = \frac{an}{(a-n)^{2}} > \frac{ab}{(a-b)^{2}} = a$, contradiction. If $(a-b)^{2} = a$, we can without loss of generality assume that $n$ is the second largest element of $A$. Then $\frac{[n,a]}{a-n} = \frac{an}{(a-n)^{2}} > \frac{an}{(a-b)^{2}} = n$ so $\frac{an}{(a-n)^{2}} = a \implies n = (a-n)^{2} \implies (a-b)^{2} = a = (a-n)^{2} + (a-n)$ and $(a-b)^{2} - (a-n)^{2} = a-n \implies (n-b)(a-b+a-n) = a-n$ which is a contradiction since $a-b,n-b > 0$. |
determinant linear factorisation
there's this determinant problem I've been working on for several days now whose answer I can't quite get to:
$$D = \left| \begin{array}{ccc} a^3+a^2 & a & 1\\ b^3+b^2 & b & 1\\ c^3+c^2 & c & 1\\ \end{array} \right|$$
Express the determinant as the product of four linear factors.
The given answer is $(a-b)(b-c)(c-a)(a+b+c+1)$ but I'm stuck after getting the first two factors, $(a-b)$ and $(b-c)$:
$$D= (a-b)(b-c) \left| \begin{array}{ccc} a^2+ab+b^2 & a-b & 0\\ b^2+bc+c^2 & b-c & 0\\ c^3+c^2 & c & 1\\ \end{array} \right|$$
whose determinant transposes into:
$$D=(a-b)(b-c) \left| \begin{array}{ccc} a^2+ab+b^2 & b^2+bc+c^2 & c^3+c^2\\ 1 & 1 & c\\ 0 & 0 & 1\\ \end {array} \right| \\ \Rightarrow D=(a-b)(b-c))1(a^2+ab+b^2)-1(b^2+bc+c^2))\\ =(a-b)(b-c)(a^2+ab+bc+c^2)$$
This is where I am stuck.
-
Playing around, and beginning pretty much as you began (except carrying out the row operations correctly): $$D = \begin{vmatrix} a^3+a^2 & a & 1 \\ b^3+b^2 & b & 1 \\ c^3+c^2 & c & 1 \end{vmatrix}\\ = \begin{vmatrix} (a^3-b^3)+(a^2-b^2) & a-b & 0 \\ (b^3-c^3)+(b^2-c^2) & b-c & 0 \\ c^3+c^2 & c & 1 \end{vmatrix}\\ = \begin{vmatrix} (a^3-b^3)+(a^2-b^2) & a-b \\ (b^3-c^3)+(b^2-c^2) & b-c \end{vmatrix}\\ = (a-b)(b-c) \begin{vmatrix} a^2+ab+b^2+a+b & 1 \\ b^2+bc+c^2+b+c & 1 \end{vmatrix}\\ = (a-b)(b-c)(a^2+ab+b^2+a+b-b^2-bc-c^2-b-c)\\ = (a-b)(b-c)(a^2+ab+a-bc-c^2-c)\\ = (a-b)(b-c)(a^2+ab+a+ac-ac-bc-c^2-c)\\ = (a-b)(b-c)(a-c)(a+b+c+1),$$ which, annoyingly enough, is minus the answer you're given.
-
Indeed, thank you! – Branimir Ćaćić Mar 18 '13 at 8:52
Hint: Sorry, you did a mistake at $D= (a-b)(b-c) \left| \begin{array}{ccc} a^2+ab+b^2 & a-b & 0\\ b^2+bc+c^2 & b-c & 0\\ c^3+c^2 & c & 1\\ \end{array} \right|$. It must be $D= (a-b)(b-c) \left| \begin{array}{ccc} a^2+ab+b^2 & 1 & 0\\ b^2+bc+c^2 & 1 & 0\\ c^3+c^2 & c & 1\\ \end{array} \right|$.
Now, what is the determinant if you expand along the third column? It is a $2\times 2$ determinant $\left| \begin{array}{cc} a^2+ab+b^2 & 1 \\ b^2+bc+c^2 & 1 \\ \end{array} \right|$.
Also, though your approach is fine, there may be alternative ways as well.
Edit: Here is some more hints to find the determinant (possibly "without expanding"):
First note that $D=D_1+D_2$, where $D_1 = \left| \begin{array}{ccc} a^3 & a & 1\\ b^3 & b & 1\\ c^3 & c & 1\\ \end{array} \right|$. Now to find the value of $D_1$, note that if we put $a=b$, or $b=c$, or $c=a$, the determinant vanishes. So, $(a-b)(b-c)(c-a)$ must be factors of the determinant. Now, note that the determinant has a leading term (=product of diagonal entries) $a^3b$. So, the other factor must be a linear...and finally the determinant is actually $-(a-b)(b-c)(c-a)(a+b+c)$. Apply the same technique to find $D_2$ and then add.
-
In any case, your transposition step is not correct. How did you get it? – Tapu Mar 18 '13 at 8:06
This also gives the same value when expanded, so I'm not too sure where to start again because I do want those $(a-b)$ and $(b-c)$ factors. – Harris Crescendo Mar 18 '13 at 8:20
Please see my Edits...for another way. – Tapu Mar 18 '13 at 8:26 |
0
Q:
# The true discount on a bill of Rs. 2160 is Rs. 360. What is the banker's discount?
A) Rs. 432 B) Rs. 422 C) Rs. 412 D) Rs. 442
Explanation:
F = Rs. 2160
TD = Rs. 360
PW = F - TD = 2160 - 360 = Rs. 1800
True Discount is the Simple Interest on the present value for unexpired time
=>Simple Interest on Rs. 1800 for unexpired time = Rs. 360
Banker's Discount is the Simple Interest on the face value of the bill for unexpired time
= Simple Interest on Rs. 2160 for unexpired time
=(360/1800)* 2160
= (1/5) * 2160
=Rs. 432
Q:
The bankers discount and the true discount of a sum at 10% per annum simple interest for the same time are Rs.100 and Rs.80 respectively. What is the sum and the time?
A) Sum = Rs.400 and Time = 5 years B) Sum = Rs.200 and Time = 2.5 years C) Sum = Rs.400 and Time = 2.5 years D) Sum = Rs.200 and Time = 5 years
Answer & Explanation Answer: C) Sum = Rs.400 and Time = 2.5 years
Explanation:
BD = Rs.100
TD = Rs.80
R = 10%
$F=\frac{BD×TD}{BD-TD}=\frac{100×80}{100-80}=Rs.400$
BD = Simple interest on the face value of the bill for unexpired time= FTR/100
$⇒100=\frac{400×T×10}{100}\phantom{\rule{0ex}{0ex}}$
=> T = 2.5 years
2 1733
Q:
If the discount on Rs. 498 at 5% simple interest is Rs.18, when is the sum due?
A) 8 months B) 11 months C) 10 months D) 9 months
Explanation:
F = Rs. 498
TD = Rs. 18
PW = F - TD = 498 - 18 = Rs. 480
R = 5%
$TD=\frac{PW×TR}{100}$
$⇒18=\frac{480×T×5}{100}$
=> T = 3/4 years = 9 months
1 1461
Q:
What is the difference between the banker's discount and the true discount on Rs.8100 for 3 months at 5%
A) Rs. 2 B) Rs. 1.25 C) Rs. 2.25 D) Rs. 0.5
Explanation:
F = Rs. 8100
R = 5%
T = 3 months = 1/4 years
$BD=\frac{FTR}{100}=\frac{8100×\frac{1}{4}×5}{100}=Rs.101.25$
$TD=\frac{FTR}{100+TR}=\frac{8100×\frac{1}{4}×5}{100+\left(\frac{1}{4}×5\right)}=Rs.100$
Therefore BD - TD = 101.25-100 = Rs.1.25
1 1897
Q:
The banker's discount on a bill due 6 months hence at 6% is Rs. 18.54. What is the true discount?
A) Rs. 24 B) Rs. 12 C) Rs. 36 D) Rs. 18
Explanation:
T= 6 months = 1/2 yearR = 6%
0 2411
Q:
The B.D. and T.D. on a certain sum is Rs.200 and Rs.100 respectively. Find out the sum.
A) Rs. 400 B) Rs. 300 C) Rs. 100 D) Rs. 200
Explanation:
$F=\frac{BD×TD}{BD-TD}=\frac{200×100}{200-100}=\frac{200×100}{100}=Rs.200$
0 1002
Q:
The B.G. on a certain sum 4 years hence at 5% is Rs. 200. What is the present worth?
A) Rs. 4500 B) Rs. 6000 C) Rs. 5000 D) Rs. 4000
Explanation:
T = 4 years
R = 5%
Banker's Gain, BG = Rs.200
$TD=\sqrt{PW×BG}$
$⇒1000=\sqrt{PW×200}$
=>PW = Rs.5000
3 1533
Q:
What is the present worth of a bill of Rs.1764 due 2 years hence at 5% compound interest is
A) Rs. 1600 B) Rs. 1200 C) Rs. 1800 D) Rs. 1400
Explanation:
Since the compound interest is taken here,
$PW{\left(1+\frac{5}{100}\right)}^{2}=1764$
=> PW = 1600
0 1030
Q:
The present worth of a certain sum due sometime hence is Rs. 3400 and the true discount is Rs. 340. The banker's gain is:
A) Rs. 21 B) Rs. 17 C) Rs. 18 D) Rs. 34
$BG=\frac{{\left(TD\right)}^{2}}{PW}=\frac{\left(340{\right)}^{2}}{3400}=Rs.34$ |
Hong Kong
Stage 4 - Stage 5
# Classifying conic sections
Lesson
## Generating a conic section
A conic section (often just called a conic) is a curve generated by the intersection of the surface of a cone with a plane.
As this diagram shows, depending on the angle that the cutting plane makes with the cone, one of four curves are obtained.
• If the plane cuts the cone horizontally, the closed curve generated becomes a circle (shown in red)
• If the cutting plane is tilted at an angle, and cuts through to make a closed curve, that curve becomes an ellipse (shown in green).
• If the cutting plane is tilted so that the major axis of the curve generated (shown as a dotted line through the blue curve) is parallel to the slant edge of the cone, then the curve becomes a parabola.
• If the cutting plane is tilted further than that required to generate a parabola, then the curve becomes a hyperbola (shown in orange).
In the case of the hyperbola, an inverted cone balanced upside with its apex touching the apex of the first cone is often shown in the diagram (the pair of cones are referred to as a double napped cone). The plane will cut both cones and generate two hyperbolic branches. These branches will be reflections of each other if the cutting plane becomes vertical.
The Greek mathematician Appolonius of Perga around 200 BCE investigated these curves extensively, but it was not until the 17th Century with the emergence of the scientific revolution that natural philosophers such as Johannes Kepler and Isaac Newton realised the physical significance of the four sections. For example, the path of a water stream from a hose is parabolic, the path of a planet is elliptical and the path of a fast comet that sweeps by our Sun once will curve around it as a hyperbola.
This neat animated applet shows the various cuts creating the different conics.
## The conics as loci
A conic section can also be established as the locus of points.
Imagine walking between a tree that lies close to a river's edge in such a way as to keep an equal distance between both:
In mathematical terms we might describe the locus as all points $P$P such that the distance from the base of the tree $S$S to $P$P is always the same distance from $P$P to the bank of the river $M$M. In other words, $SP=PM$SP=PM
Remarkably, we can show that this path is a parabola.
If we changed the locus so that $SP=2PM$SP=2PM (a path closer to the river than the tree), the path changes to a hyperbola. In fact this will happen if $SP=ePM$SP=ePM where $e>1$e>1.
If we changed the locus so that $SP=SP=\frac{1}{2}PM$SP=SP=12PM (a path closer to the tree than the river) the path becomes an ellipse - a closed loop around the tree! Again, this will always happen if $SP=ePM$SP=ePM and $e<1$e<1, shown here:
## Overlaying a coordinate system
To discover the many properties of conic sections, it is convenient to overlay a cartesian coordinate system in such a way as to simplify the mathematics that arises.
To this end, we place the coordinate axes so as to make the lines of symmetry of the conic parallel and perpendicular to these axes, as shown here using three examples:
For the ellipse and the hyperbola, the red dot in the picture represents the centre of the conic, and in the parabola the red dot locates its vertex.
## The equation of a conic
Now when we set up the axes in this way, the equation of every conic (including the circle) has the form:
$Ax^2+By^2+Cx+Dy+E=0$Ax2+By2+Cx+Dy+E=0
with $A,B,C,D$A,B,C,D and $E$E are coefficients of each of the terms in the equation.
To be a conic, either $A$A or $B$B or both must be non-zero (note that if $A=B=0$A=B=0 the conic would degenerate into the straight line $Cx+Dy+E=0$Cx+Dy+E=0).
In fact we can classify the type of conic based on $A$A and $B$B:
• If the conic is an ellipse, then $AB$AB is positive
• An ellipse that has $A=B$A=B is a circle
• If the conic is a hyperbola, then $AB$AB is negative
• If the conic is a parabola, then either $A$A or $B$B (but not both) is zero
Unfortunately, these statements don't necessarily work the other way around (See example 4 below).
If either $C$C or $D$D or both are non-zero, then the centre of the hyperbola or ellipse will be located away from the origin. The same applies to the parabola - if $C$C or $D$D or both are non-zero, the vertex will be away from the origin.
As for the constant coefficient $E$E, there are certain circumstances where degenerate forms can arise but these are best understood by considering a few examples.
With the axes set up like this, the equation of the hyperbola and the ellipse can always be rearranged into the standard forms $\frac{\left(x-h\right)^2}{a^2}-\frac{\left(y-k\right)^2}{b^2}=\pm1$(xh)2a2(yk)2b2=±1 and $\frac{\left(x-h\right)^2}{a^2}+\frac{\left(y-k\right)^2}{b^2}=1$(xh)2a2+(yk)2b2=1 respectively where $\left(h,k\right)$(h,k) locates the centre.
Likewise, the equation of the parabola can always be rearranged into the vertex form $y-k=a\left(x-h\right)^2$yk=a(xh)2 where the vertex is located at $\left(h,k\right)$(h,k).
### Some examples
##### Example 1
What type of conic is given by the equation $x^2+y^2-6x+4y-12=0$x2+y26x+4y12=0. Where is the conic's centre?
Since $A=B=1$A=B=1, the conic is a circle. By completing squares on $x$x and $y$y, we can locate the centre:
$x^2+y^2-6x+4y-12$x2+y2−6x+4y−12 $=$= $0$0 $x^2-6x+y^2+4y$x2−6x+y2+4y $=$= $12$12 $x^2-6x+9+y^2+4y+4$x2−6x+9+y2+4y+4 $=$= $12+9+4$12+9+4 $\left(x-3\right)^2+\left(y+2\right)^2$(x−3)2+(y+2)2 $=$= $25$25
This means that the circle's centre is located at $\left(3,-2\right)$(3,2) and the circle's radius is $5$5.
##### Example 2
What type of conic is given by the equation $x^2-6x-3y-12=0$x26x3y12=0. Where is the conic's centre?
Since $B=0$B=0, the equation is that for a parabola. Again, completing squares:
$x^2-6x-3y-12$x2−6x−3y−12 $=$= $0$0 $x^2-6x+9-3y-21$x2−6x+9−3y−21 $=$= $0$0 $\left(x-3\right)^2-3\left(y+7\right)$(x−3)2−3(y+7) $=$= $0$0 $y+7$y+7 $=$= $\frac{1}{3}\left(x-3\right)^2$13(x−3)2
Hence the parabola opens upward with vertex $\left(3,-7\right)$(3,7)
##### Example 3
What type of conic is given by the equation $9x^2+4y^2+18x-16y-11=0$9x2+4y2+18x16y11=0? Where is the conic's centre, and find its $x$x intercepts?
Since $A\ne B$AB and $AB$AB is positive, the equation represents an ellipse.
The $x$x intercepts are found by setting $y=0$y=0 and thus solving for $x$x in $9x^2+18x-11=0$9x2+18x11=0. By using the quadratic formula, we find that $x=-1\pm\frac{2\sqrt{5}}{3}$x=1±253, and these are the $x$x intercepts.
To find the centre we again complete squares:
$9x^2+4y^2+18x-16y-11$9x2+4y2+18x−16y−11 $=$= $0$0 $9x^2+18x+4y^2-16y$9x2+18x+4y2−16y $=$= $11$11 $9\left(x^2+2x+1\right)+4\left(y^2-4y+4\right)$9(x2+2x+1)+4(y2−4y+4) $=$= $36$36 $9\left(x+1\right)^2+4\left(y-2\right)^2$9(x+1)2+4(y−2)2 $=$= $36$36 $\frac{\left(x+1\right)^2}{4}+\frac{\left(y-2\right)^2}{9}$(x+1)24+(y−2)29 $=$= $1$1
Hence the centre of the ellipse is located at $\left(-1,2\right)$(1,2)
##### Example 4
Is $2x^2-3y^2-12x+30y-57=0$2x23y212x+30y57=0 a hyperbola?
Since $AB=2\times-3=-6$AB=2×3=6, the conic looks to be a hyperbola. However, when we complete squares, we find that the equation becomes $2\left(x-3\right)^2-3\left(y-5\right)^2=0$2(x3)23(y5)2=0.
Without going into the algebra, the left hand side can be factorised into the difference of two squares and each factor can be brought to zero, so that the equation represents two straight lines with equations given by $\sqrt{2}\left(x-3\right)-\sqrt{3}\left(y-5\right)=0$2(x3)3(y5)=0 and $\sqrt{2}\left(x-3\right)+\sqrt{3}\left(y-5\right)=0$2(x3)+3(y5)=0 as shown here:
Thus it is a necessary, but not sufficient condition that $AB$AB to be negative for the equation to represent a hyperbola.
#### Mathspace Worked Examples
##### Question 1
What is the graph of a conic section represented by the equation $x^2-10x-y=0$x210xy=0?
1. Parabola
A
Ellipse
B
Hyperbola
C
Circle
D
##### Question 2
What type of graph does the equation $4x^2+32x-9y^2+54y-53=0$4x2+32x9y2+54y53=0 represent?
1. Circle
A
Ellipse
B
Parabola
C
Hyperbola
D
##### Question 3
What type of graph does the conic section represented by the equation $\frac{x^2}{9}+\frac{y^2}{4}=1$x29+y24=1 have?
1. Hyperbola
A
Ellipse
B
Parabola
C
Circle
D |
3.09 Graphs of linear inequalities
Lesson
Previously, we were introduced to the four inequality symbols, and learnt how to solve simple inequalities. Here are some examples:
$x<2$x<2 "$x$x is less than $2$2" $x>-5$x>−5 "$x$x is greater than $-5$−5" $2x\le-4$2x≤−4 "$2$2 groups of $x$x is less than or equal to $-4$−4" $x-3\ge17$x−3≥17 "$3$3 less than $x$x is greater than or equal to $17$17"
Inequalities that include a variable, such as the examples above, can be represented nicely on a number line. Let's quickly recap plotting points on a number line as learnt in Grade 7.
The Number Line
Remember that all the real numbers can be represented on an infinite line called the number line, stretching off towards positive infinity on the right, and negative infinity on the left. Numbers further to the left are smaller numbers and numbers further to the right are larger numbers.
We can plot any real number we like on the number line. For example, if we know that $x=6$x=6, we can plot the value of $x$x as with a solid dot:
A plot of $x=6$x=6.
Similarly, if we know that $x=\frac{19}{5}$x=195, we can plot the value of $x=3\frac{4}{5}$x=345 as follows:
A plot of $x=\frac{19}{5}$x=195.
Inequalities on the Number Line
Now, what if we wanted to plot an inequality, such as $x\le4$x4? We can review this from Grade 7.
When we say "$x$x is less than or equal to $4$4", we're not just talking about one number. We're talking about a whole set of numbers, including $x=4$x=4, $x=2$x=2, $x=0$x=0, $x=-1$x=1 and $x=-1000$x=1000. All of these numbers are less than or equal to $4$4.
If we plot all of the integers that are less than or equal to $4$4 on a number line, we get something that looks like this:
A first attempt at plotting $x\le4$x4.
So far so good. But what about fractions like $x=\frac{1}{2}$x=12, or irrational numbers like $x=\sqrt{2}$x=2?
These numbers are also less than or equal to $4$4, so surely they should be shown on the plot too?
To show all of the values less than or equal to $4$4, we can draw a ray (a directed line) to represent all of these points, since all of them are included in the inequality.
The actual plot of $x\le4$x4.
What if we instead want to plot the very similar inequality $x<4$x<4? The only difference now is that $x$x cannot take the value of $4$4, and so the plot should not include the point where $x=4$x=4.
So we want to plot the same ray, but leave off the point at the end where $x=4$x=4. To represent this we draw the plot with a hollow circle, instead of a filled in circle, to show that $4$4 is not included:
A plot of $x<4$x<4.
To plot a greater than or greater than or equal to inequality, we instead want to show all of the numbers with larger value than a particular number. This is as easy as drawing a ray in the other direction instead, pointing to the right off towards positive infinity. For example, the inequalities $x\ge4$x4 and $x>4$x>4 are plotted below:
A plot of $x\ge4$x4.
A plot of $x>4$x>4.
Here are examples of inequalities plotted on a number line that have negative values:
A plot of $x>-\frac{19}{20}$x>1920.
A plot of $x\le-20.7$x20.7.
Remember!
• For $\le$ and $\ge$ we use a filled in or closed dot to start the ray, to show the starting point is included
• For $<$< and $>$> we use a hollow or open dot to start the ray, to show the starting point is not included
• To check your ray is going the right way, choose a value which satisfies the inequality and make sure your ray covers it
Practice Questions
Question 1
State the inequality for $x$x that is represented on the number line.
Question 2
Plot the inequality $x\le1$x1 on the number line below.
Question 3
Consider the inequality $-3x+7\ge4$3x+74.
1. Solve the inequality.
2. Now plot the solutions to the inequality $-3x+7\ge4$3x+74 on the number line below.
Outcomes
8.C2.4
Solve inequalities that involve integers, and verify and graph the solutions. |
# NCERT Solutions for Class 8 Maths Chapter 10 Visualising Solid Shapes Ex 10.3
NCERT Solutions for Class 8 Maths Chapter 10 Visualising Solid Shapes Ex 10.3
### NCERT Solutions for Class 8 Maths Chapter 10 Visualising Solid Shapes Exercise 10.3
Ex 10.3 Class 8 Maths Question 1.
Can a polyhedron have for its faces
(i) 3 triangles?
(ii) 4 triangles?
(iii) a square and four triangles?
Solution:
(i) No, because polyhedron must have edges meeting at vertices which are points.
(ii) Yes, because all the edges are meeting at the vertices.
(iii) Yes, because all the eight edges meet at the vertices.
Ex 10.3 Class 8 Maths Question 2.
Is it possible to have a polyhedron with any given number of faces?
(Hint: Think of a pyramid)
Solution:
Yes, it is possible if the number of faces is greater than or equal to 4.
Example: Pyramid which has 4 faces.
Ex 10.3 Class 8 Maths Question 3.
Which are prisms among the following?
Solution:
Only (ii) unsharpened pencil and (iv) a box are the prism.
Ex 10.3 Class 8 Maths Question 4.
(i) How are prisms and cylinders alike?
(ii) How are pyramids and cones alike?
Solution:
(i) If the number of sides in a prism is increased to certain extent, then the prism will take the shape of cylinder.
(ii) If the number of sides of the pyramid is increased to same extent, then the pyramid becomes a cone.
Ex 10.3 Class 8 Maths Question 5.
Is a square prism same as a cube? Explain.
Solution:
Every square prism cannot be cube. It may be cuboid also.
Ex 10.3 Class 8 Maths Question 6.
Verify Euler’s formula for these solids.
Solution:
(i) Faces = 7
Sides = 15
Vertices = 10
Euler’s formula: F + V – E = 2
⇒ 7 + 10 – 15 = 2
⇒ 2 = 2
Hence, Euler’s formula is verified.
(ii) Faces = 9
Sides = 16
Vertices = 9
Euler’s Formula: F + V – E = 2
⇒ 9 + 9 – 16 = 2
⇒ 2 = 2
Hence, Euler’s formula is verified.
Ex 10.3 Class 8 Maths Question 7.
Using Euler’s formula find the unknown.
Faces ? 5 20 Vertices 6 ? 12 Edges 12 9 ?
Solution:
Faces 8 5 20 Vertices 6 6 12 Edges 12 9 30
Using Eulers Formula: F + V – E = 2
Ex 10.3 Class 8 Maths Question 8.
Can a polyhedron have 10 faces, 20 edges and 15 vertices?
Solution:
Here faces = 10, Edges = 20, Vertices = 15
According to Euler’s Formula:
F + V – E = 2
⇒ 10 + 15 – 20 = 25 – 20
⇒ 5 ≠ 2
A polyhedron do not have 10 Faces, 20 Edges and 15 Vertices.
## SabDekho
The Complete Educational Website |
# Algebra I CM third marking term
## Presentation on theme: "Algebra I CM third marking term"— Presentation transcript:
Algebra I CM third marking term
Wicomico High School Mrs. J. Austin Chapter 7 : System of Equations Welcome to Algebra CM minutes Seating Chart Classroom Sets of Books
Systems of Linear Equations
Two linear equations graphed on the SAME coordinate plane. What are the THREE things that could happen? They could CROSS or INTERSECT They could NEVER CROSS or be PARALLEL They could be ON TOP OF EACH OTHER or COINCIDE
Solving a System of Linear Equations
By Solving a System of Linear Equations, we are asking: Are there any Values for x and y that will “satisfy” or make BOTH equations TRUE? Is there a POINT that will make BOTH equations TRUE? What is the POINT OF INTERSECTION of these two lines? Find the values for x and y that will make BOTH equations TRUE.
Solving a System By Graphing 7.1
Transform each equation to Slope-Intercept Form For EACH of the TWO equations: PLOT the y – intercept, b COUNT, rise over run using the Slope, m. DRAW the straight line. Cognitive Tutor Packet: Intro to Systems
Solving a System By Graphing 7.1
Transform each equation into Slope-Intercept Form.
Solving a System By Substitution 7.2
Transitive Property A variable can be REPLACED with its equivalent. If two equations equal the SAME thing, they must then EQUAL each other. AND THEN The two equations are SET equal to each other Textbook Use Cognitive Tutor Packet #1
Solving a System By Substitution
Solve the first equation for x. Substitute the expression in for x in the second equation. Now SOLVE for y. EXAMPLE: Now we know y=2. Substitute this into an equations to find x.
Solving a System By Elimination 7.3
Transitive Property Replace a variable with its equivalent. Set two expressions equal to each other, when they BOTH equal the same thing! Property of Equality Add or Subtract two equations to create a new equivalent equation. Transform the look of an equation by multiplication. Cognitive Tutor Packet #2
Solving a System By Elimination 7.3
Using Addition: ___________ Write the equations one above the other. Be sure the variables are lined up. Draw a line under them. Combine the Like-Terms to create a NEW equation. Solve for the variable. Substitute your answer into one of the equations to find the other variable.
Solving a System By Elimination 7.3
Write the equations one above the other. Be sure the variables are lined up. Draw a line under them. Subtract the Like-Terms to create a NEW equation. Solve for the variable. Substitute your answer into one of the equations to find the other variable. Using Subtraction _____________ Cognitive Tutor Packet #3
Solving a System By Elimination 7.4
Using Multiplication Before Adding or Subtracting: ___________________ __________________ Write the equations one above the other. Be sure the variables are lined up. Multiply the top equation by 4. Multiply the lower equation by 7. Draw a line under them. Distribute through each equation. Combine or Subtract the Like-Terms to create a NEW equation. Cognitive Tutor Packet #4
Solving a System By Elimination 7.4
Using Multiplication Before Adding or Subtracting What would you MULTIPLY by? Multiply the top equation by 3. Solve the System using Subtraction: ____________________ The solution is the Point of Intersection: Solve this System. Start by multiplying the top equation by 2. Then add the two equations together eliminating the x terms. Solve for y. Substitute in the y value and solve for x. The solution is: . Cognitive Tutor Packet #4
Special Types of Systems 7.5
One Solution: Lines Intersect at one point. Lines have different slopes. Lines may be PERPENDICULAR if they cross at 90⁰ angles. No Solution: Lines do not intersect. Lines have the SAME slope and DIFFERENT y-intercepts. Lines are PARALLEL. Many Solutions: Lines touch on every point. Lines have the SAME slope and SAME y-intercepts. Line COINCIDE.
Special Types of Systems 7.5
How Many Solutions Does the System Have? . System: Answer Choices: One Intersecting Lines None Parallel Lines Many Coinciding Lines
Writing and Solving Systems
Slope –Intercept Form: Total Cost scenarios with given rates of change. The movie theater charges \$8 per a ticket for its general customers. It offers a movie club discount of \$5 per ticket if you join the club for a one-time fee of \$15. How many movies would you have to go see to make joining the club beneficial? Write a system. Let x = the number of movie tickets y = total cost Mixed Review: Pg. 465 (42-61)
Writing and Solving Systems
Standard or General Form: Two different Items are given. At a grocery store, a customer pays a total of \$9.70 for 1.8 pounds of potato salad and 1.4 pounds of coleslaw. Another customer pays a total of \$6.55 for 1 pound of potato salad and 1.2 pounds of coleslaw. How much do 2 pounds of potato salad and 2 pounds of coleslaw cost? Write the system. Let: x = cost of potato salad y = cost of coleslaw. Mixed Review: Pg. 465 (42-61)
Writing and Solving Systems
High School Assessment Practice Questions: READ the question ALL the way through. RE-READ and define the variables. RE-READ and WRITE two equations to model the scenario. DECIDE which METHOD you will use to solve the System of Equations. SOLVE the System. RE-READ the question. Use YOUR SOLUTION to CONSTRUCT a written ANSWER to the question. HSA Packet
Solving Systems of Linear Inequalities 7.6
Graphing Linear Inequalities: Graph the first line. Shade the area defined by the first line. Graph the second line. Shade the area defined by the second line. The SOLUTUION to the System of Linear Inequalities is the AREA OF INTERSECTION. Re-shade the section of the graph that has been shaded by both of the equations.
Chp 7 Review Two linear functions graphed on a coordinate plane.
They could CROSS, INTERSECT They could NEVER CROSS, PARALLEL They could be ON TOP OF EACH OTHER, COINCIDE Solving a System By Graphing 7.1 Solving a System By Substitution 7.2 Solving a System By Elimination 7.3 Identifying the Point of Intersection Testing a Solution to a System of Equations Special Types of Systems Writing and Solving Systems of Equations |
# 10.3 Multiply Polynomials
### Learning Objectives
By the end of this section, you will be able to:
• Multiply a polynomial by a monomial
• Multiply a binomial by a binomial
• Multiply a trinomial by a binomial
#### Multiply a Polynomial by a Monomial
In Distributive Property, you learned to use the Distributive Property to simplify expressions such as 2(x−3). You multiplied both terms in the parentheses, xand3, by 2, to get 2x−6. With this chapter’s new vocabulary, you can say you were multiplying a binomial, x−3, by a monomial, 2. Multiplying a binomial by a monomial is nothing new for you!
Multiplying a monomial by a trinomial works in much the same way.
Now we will have the monomial as the second factor.
### Multiply a Binomial by a Binomial
Just like there are different ways to represent multiplication of numbers, there are several methods that can be used to multiply a binomial times a binomial.
#### Using the Distributive Property
We will start by using the Distributive Property. Look again at Example 10.33.
Notice that before combining like terms, we had four terms. We multiplied the two terms of the first binomial by the two terms of the second binomial—four multiplications.
Be careful to distinguish between a sum and a product.
Now we’ll see how to multiply binomials where the variable has a coefficient.
In the previous examples, the binomials were sums. When there are differences, we pay special attention to make sure the signs of the product are correct.
Up to this point, the product of two binomials has been a trinomial. This is not always the case.
#### Using the FOIL Method
Remember that when you multiply a binomial by a binomial you get four terms. Sometimes you can combine like terms to get a trinomial, but sometimes there are no like terms to combine. Let’s look at the last example again and pay particular attention to how we got the four terms.
Where did the first term, x2,come from?
It is the product of xandx, the first terms in (x+2)and(xy).
The next term, −xy, is the product of xand−y,
the two outer terms.
The third term, +2x, is the product of 2andx,
the two inner terms.
And the last term, −2y,
came from multiplying the two last terms.
We abbreviate “First, Outer, Inner, Last” as FOIL. The letters stand for ‘First, Outer, Inner, Last’. The word FOIL is easy to remember and ensures we find all four products. We might say we use the FOIL method to multiply two binomials.
Let’s look at (x+3)(x+7)
again. Now we will work through an example where we use the FOIL pattern to multiply two binomials.
We summarize the steps of the FOIL method below. The FOIL method only applies to multiplying binomials, not other polynomials!
#### Using the Vertical Method
The FOIL method is usually the quickest method for multiplying two binomials, but it works only for binomials. You can use the Distributive Property to find the product of any two polynomials. Another method that works for all polynomials is the Vertical Method. It is very much like the method you use to multiply whole numbers. Look carefully at this example of multiplying two-digit numbers.
You start by multiplying 23 by 6 to get 138.
Then you multiply 23 by 4, lining up the partial product in the correct columns.
Last, you add the partial products.
Now we’ll apply this same method to multiply two binomials.
We have now used three methods for multiplying binomials. Be sure to practice each method, and try to decide which one you prefer. The three methods are listed here to help you remember them.
### Multiplying Two Binomials
To multiply binomials, use the:
• Distributive Property
• FOIL Method
• Vertical Method
Remember, FOIL only works when multiplying two binomials.
### Multiply a Trinomial by a Binomial
We have multiplied monomials by monomials, monomials by polynomials, and binomials by binomials. Now we’re ready to multiply a trinomial by a binomial. Remember, the FOIL method will not work in this case, but we can use either the Distributive Property or the Vertical Method. We first look at an example using the Distributive Property.
Now let’s do this same multiplication using the Vertical Method.
#### Media
• Multiply Monomials
• Multiply Polynomials
• Multiply Polynomials 2
• Multiply Polynomials Review
• Multiply Polynomials Using the Distributive Property
• Multiply Binomials
#### Practice Makes Perfect
Multiply a Polynomial by a Monomial
In the following exercises, multiply.
Multiply a Binomial by a Binomial
In the following exercises, multiply the following binomials using: ⓐ the Distributive Property ⓑ the FOIL method ⓒ the Vertical method
Multiply a Trinomial by a Binomial
In the following exercises, multiply using ⓐ the Distributive Property and ⓑ the Vertical Method.
#### Self Check
ⓐ After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.
ⓑ What does this checklist tell you about your mastery of this section? What steps will you take to improve? |
# SOCR EduMaterials Activities JointDistributions
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)
## This is an activity to explore he joint distributions of X and Y through two simple examples.
You can access the following experiments in SOCR under Experiments.
• Exercise 1: Die coin experiment:
A die is rolled and the number observed \( X \) is recorded. Then a coin is tossed number of times equal to the value of \(X</math?. For example if [itex]X=2\) then the coin is tossed twice, etc. Let \(Y\) be the number of heads observed. Note: Assume that the die and the coin are fair.
• 1. Construct the joint probability distribution of \(X\) and \(Y\).
• 2. Find the conditional expected value of \(Y\) given \(X=5\).
• 3. Find the conditional variance of \(Y\) given \(X=5\).
• 4. Find the expected value of \(Y\).
• 5. Find the standard deviation of \(Y\).
• 6. Graph the probability distribution of \(Y\).
• 7. Use SOCR to graph and print the empirical distribution of \(Y\) when the experiment is performed
• a. \(n=1000\) times.
• b. \(n=10000\) times.
• 8. Compare the theoretical mean and standard deviation of \(Y\) (parts (4) and (5)) with the empirical mean and standard deviation found in part (8).
Below you can see a snapshot of the theoretical distribution of \( Y \).
• Exercise 2: Coin Die experiment:
A coin is tossed and if heads is observed then a red die is rolled. If tails is observed then a green die is rolled. You can choose the distribution of each die as well as the probability of heads. Choose for the red die the 3-4 flat distribution and for the green die the skewed right distribution. Finally using the scroll button choose \$p=0.2\$ as the probability of heads. Let \(X\) be the score of the coin (1 for heads, 0 for tails), and let \(Y\) be the score of the die (1,2,3,4,5,6).
• 1. Construct the joint probability distribution of \(X, Y\).
• 2. Find the marginal probability distribution of \(Y\) and verify that it is the same with the one given in the applet.
• 3. Compute \(E(Y)\).
• 4. Compute \(E(Y)\) using expectation by conditioning \(E[E(Y|X)]\).
• 5. Run the experiment 1000 times take a snapshot and comment on the results.
Below you can see a snapshot of the theoretical distribution of \( Y \).
{{translate|pageName=http://wiki.stat.ucla.edu/socr/index.php?title=SOCR_EduMaterials_Activities} |
# Find the 1 Bogus Coin in 50 - and Fast
RAY: Here's how:
Step 1: Divide the coins into three piles. A pile of 17, another pile of 17 and a pile of 16.
Step 2: Weigh the two piles of 17 on your balance beam. This is your first weighing.
For the moment, let's assume that one pile is heavier. We know that the bogus coin is in that group. In that case, we need to take the following steps:
Divide the heavier pile of 17 into a pile of 6, another pile of 6 and a pile of 5.
Weigh the two piles of 6. Let's assume that one pile is heavier. In this case, take the heavier pile of six and break it into three piles of two. For your third weighing, weigh two of the piles. If one pile is heavier, weigh those two coins against each other to find the bogus coin. If not, weigh the pile of two that you put aside to find the heavier coin.
Let's go back a step and see what happens if those two piles of six you weighed were equal. In this case, you know that the heavy coin is in the group of 5 coins. So, take the pile of 5 and break it into a pile of 2, a pile of 2 and 1 coin. Weigh the 2 piles of 2 against each other. If they're equal, you know the bogus coin is the one you didn't weigh. If they're not equal, take the heavier pile of 2, split it and weigh those 2 coins to find the bogus one.
Now, let's go all the way back to Step 2. Let's now assume that two piles of 17 coins balanced when you weighed them and the heavy coin was in the group of 16. In that case, you'd take that pile of 16 and break it into a pile of 5, another pile of 5 and a pile of 6, and your next step would be to weigh the 2 piles of 5 against each other. There are two possible outcomes here:
If the 2 piles of 5 are equal, you know the heavy coin is in the pile of 6. In this case, break the 6 coins in to 3 piles of 2. Weigh 2 piles against each other. If they're equal, you know the bogus coin is in the pile you didn't weigh, and you can find the coin in one more weighing. If they're not equal, take the heavier pile of 2, split it and weigh those 2 coins to find the bogus one.
If the 2 piles of 5 are unequal, you know the bogus coin is in the heavier pile. In that case, take the heavier pile and divide it into a pile of 2, another pile of 2 and a pile of just 1. If you can't figure out the bogus coin from here in no more than two weighs, you're in big trouble!
Pretty neat, huh?
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# Points (5 ,4 ) and (2 ,2 ) are (5 pi)/4 radians apart on a circle. What is the shortest arc length between the points?
Feb 11, 2018
the shortest arc happens to be $s = 4.957$
#### Explanation:
Given:
$\left(5 , 4\right)$ is a point on the circle
$\left(2 , 2\right)$is a point on the circle
Angle subtended at the center is $\frac{5 \pi}{4}$ which is reflex lying on the major arc
The minor arc happens to be $2 \pi - \frac{5 \pi}{4} = \frac{3 \pi}{4}$
The triangle formed by the two points and the center happens to be an isoceles triangle with the shortest angle formed at the center being $\alpha = \frac{3 \pi}{4}$.
Remaining angle is
$\left(\pi - \frac{3 \pi}{4}\right) = \frac{\pi}{4}$
Angle at each of the vertex at the base is $\theta = \frac{1}{2} \frac{\pi}{4} = \frac{\pi}{8}$
Mid point happens to be
$\left(\frac{5 + 2}{2} , \frac{4 + 2}{2}\right) = \left(3.5 , 3\right)$
We calculate the radius of circle by pythagoras theorem in the right angled triangle formed by one of the point, say $\left(5 , 4\right)$, the mid-point $\left(3.5 , 3\right)$, and the center of the circle.
Distance between one of the points, say $\left(5 , 4\right)$ and the mid point $\left(3.5 , 3\right)$ is given by
${a}^{2} = {\left(3.5 - 5\right)}^{2} + {\left(3 - 4\right)}^{2} = {\left(- 1.5\right)}^{2} + {\left(- 1\right)}^{2}$
$= 2.25 + 1 = 3.25$
${a}^{2} = 3.25$
Also, the line joining $\left(5 , 4\right) \mathmr{and} \left(3.5 , 3\right)$ happens to be the adjacent side and the radius being hypotenuse for the angle at the point $\left(5 , 4\right)$
Consideing the ratio $\cos \theta$,
${\cos}^{2} \theta = {a}^{2} / {r}^{2}$
$\theta = \frac{\pi}{8}$
${a}^{2} = 3.25$
${\cos}^{2} \left(\frac{\pi}{8}\right) = \frac{3.25}{r} ^ 2$
${\cos}^{2} \frac{\pi}{8} = 0.146$
$0.854 = \frac{3.25}{r} ^ 2$
${r}^{2} = \frac{3.25}{0.854} = 3.808$
$r = \sqrt{3.808}$
$r = 1.951$
Hence, the shortest arc happens to be
$s = r \alpha = 1.951 \frac{3 \pi}{4}$
$s = 1.533$
the shortest arc happens to be $s = 4.957$
Feb 11, 2018
Shortest Arc length $s = 4.5946$
#### Explanation:
vec(AD) = sqrt(5-2)^2 + (4-2)^2) = 3.6056
$r = \frac{A D}{2 \sin \left(\frac{\theta}{2}\right)} = \frac{3.6056}{2 \sin \left(\frac{5 \pi}{8}\right)} = 1.95$
Since the center angle $\theta$ is more than $\pi$, to get the shortest arc length we must subtract from $\left(2 \pi\right)$
Shortest length of the arc
$s = r \cdot \theta = 1.95 \cdot \left(2 \pi - \left(\frac{5 \pi}{4}\right)\right) = 1.95 \cdot \left(\frac{3 \pi}{4}\right) = 4.5946$
Feb 11, 2018
Shorter arc length between the points is $4.59$ unit.
#### Explanation:
Angle subtended at the center by the arc is ${\theta}_{1} = \frac{5 \pi}{4}$
Angle subtended at the center by the minor arc is
$\theta = 2 \pi - \frac{5 \pi}{4} = \frac{3 \pi}{4} \therefore \frac{\theta}{2} = \frac{3 \pi}{8}$
Distance between two points $\left(5 , 4\right) \mathmr{and} \left(2 , 2\right)$ is
D= sqrt ((x_1-x_2)^2+(y_1-y_2)^2) =sqrt ((5-2)^2+(4-2)^2 or
$D = \sqrt{13} \approx 3.61$ unit. Chord length $l \approx 3.61$ unit.
Formula for the length of a chord is ${L}_{c} = 2 r \sin \left(\frac{\theta}{2}\right)$
where $r$ is the radius of the circle and $\theta$ is the angle
subtended at the center by the chord.
:. 3.61=2*r*sin((3pi)/8) or r = 3.61/(2*sin((3pi)/8) or
$r = \frac{3.61}{1.85} \approx 1.95 \therefore$ Radius of the circle is $1.95$ unit.
Arc length is ${L}_{a} = \cancel{2 \pi} \cdot r \cdot \frac{\theta}{\cancel{2 \pi}} = r \cdot \theta$ or
${L}_{a} = 1.95 \cdot \frac{3 \pi}{4} \approx 4.59$ unit .
Shorter arc length between the points is $4.59$ unit.[Ans] |
Reduce / leveling any portion to that lowest terms by utilizing our fraction to the Simplest form Calculator.
### Fractions Simplifier
Please fill in the two left box below:
Inputfraction: Integerpart: Fractionpart: As a Decimal: = = Details:Details...You are watching: 7/15 in simplest form
## How to minimize a fraction
Among various ways simple a fraction, us will show the two procedure below:
### Method 1 - division by a tiny Number as soon as Possible
Start by dividing both the numerator and also the denomiator of the fraction by the very same number, and repeat this until it is difficult to divide. Begin dividing by tiny numbers prefer 2, 3, 5, 7. For example,
### Simplify the portion 42/98
First divide both (numerator/denominator) by 2 to gain 21/49.Dividing by 3 and 5 will not work, so,Divide both numerator and denominator by 7 to obtain 3/7. Note: 21 ÷ 7 = 3 and also 49 ÷ 7 = 7
In the fraction 3/7, 3 is just divisible by itself, and also 7 is not divisible by various other numbers than itself and also 1, therefore the fraction has been streamlined as lot as possible. No more reduction is possible, for this reason 42/98 is same to 3/7 when lessened to its lowest terms. This is a PROPER portion once the absolute value of the peak number or molecule (3) is smaller than the absolute worth of the bottom number or denomintor (7).
### Method 2 - Greatest usual Divisor
To alleviate a portion to lowest terms (also called its simplest form), just divide both the numerator and also denominator by the GCD (Greatest common Divisor).
For example, 3/4 is in shortest form, however 6/8 is no in lowest type (the GCD the 6 and 8 is 2) and 6/8 can be written as 3/4. You can do this since the worth of a fraction will remain the same when both the numerator and also denominator are split by the same number.
Note: The Greatest common Factor (GCF) because that 6 and also 8, notation gcf(6,8), is 2. Explanation:
Factors the 6 room 1,2,3,6;Factors of 8 space 1,2,4,8.
See more: How To Apply For Nba Floor Cleaner Application {June} Read In Detail!
So, it is ease check out that the "Greatest usual Factor" or "Divisor" is 2 because it is the best number i m sorry divides same into every one of them. |
# 5Th grade order of operation: What is Order Of Operations? Definition, Rules, Examples, Facts
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## What is Order Of Operations? Definition, Rules, Examples, Facts
There are many operations in mathematics, such as addition, subtraction, multiplication, and division. They help us evaluate mathematical expressions.
Consider the following expression: 4+ 5 × 32 – 2
The expression consists of many operations. But which part do you calculate first?
You may start from the left and get one answer. But your friend may begin from the right and get a completely different answer!
Note: Both the methods given above are incorrect.
Hence, to avoid confusion, a standard rule was set to perform such calculations. This rule is known as the order of operations.
### What Is the Order of Operations in Math?
If you have an expression where all the operations are the same (example: only addition, only subtraction, only multiplication, or only division) then the correct way to solve it would be from left to right. But for expressions with multiple operations, we need to follow the order of operations.
The order of operations is the rule that tells us the sequence in which we should solve an expression with multiple operations.
A way to remember that order is PEMDAS. Each letter in PEMDAS stands for a mathematical operation.
##### Related Games
Order of Operations Steps:
#### Parentheses
The first step is to solve the operation within parentheses or brackets. Parentheses are used to group things together. Work out all groupings from inside to out.
#### Exponents
Work out the exponential expressions after the parentheses.
#### Multiplication and Division
Next, moving from left to right, multiply and/or divide, whichever comes first.
Lastly, moving from left to right, add and/or subtract, whichever comes first.
##### Related Worksheets
Why Follow the Order of Operations?
We follow the rules of the order of operations to solve expressions so that everyone arrives at the same answer.
Here’s an example of how we can get different answers if the correct order of operations is NOT followed:
### Solved Examples On Order Of Operations
Example 1: Solve: 2 + 6 × (4 + 5) ÷ 3 5 using PEMDAS.
Solution:
Step 1 – Parentheses : 2+6 × (4 + 5) ÷ 3 – 5 = 2 + 6 × 9 ÷ 3 – 5
Step 2 – Multiplication: 2 + 6 × 9 ÷ 3 – 5 = 2 + 54 ÷ 3 – 5
Step 3 – Division: 2 + 54 ÷ 3 – 5 = 2 + 18 – 5
Step 4 – Addition: 2 + 18 – 5 = 20 – 5
Step 5 – Subtraction: 20 5 = 15
Example 2: Solve 4 – 5 ÷ (8 – 3) × 2 + 5 using PEMDAS.
Solution:
Step 1 – Parentheses: 4 – 5 ÷ (8 – 3) × 2 + 5 = 4 – 5 ÷ 5 × 2 + 5
Step 2 – Division: 4 – 5 ÷ 5 × 2 + 5 = 4 – 1 × 2 + 5
Step 3 – Multiplication: 4 – 1 × 2 + 5 = 4 – 2 + 5
Step 4 – Subtraction: 4 – 2 + 5 = 2 + 5
Step 5 – Addition: 2 + 5 = 7
Example 3: Solve 100 ÷ (6 + 7 × 2) 5 using PEMDAS.
Solution:
Step 1 – Multiplication inside parentheses: 100 ÷ (6 + 7 × 2) – 5= 100 ÷ (6 + 14) – 5
Step 2 – Addition inside parentheses: 100 ÷ (6 + 14) – 5 = 100 ÷ 20 – 5
Step 3 – Division: 100 ÷ 20 – 5 = 5 – 5 Step 4 – Subtraction: 5 5 = 0
### Practice Problems On Order Of Operations
1
48
49
50
51
4 + (5 × 3² + 2)
= 4 + (5 × 9 + 2
)
= 4 + (45 + 2)
= 4 + 47
= 51
2
#### Simplify 9 – 24 ÷ 8 × 2 + 3 using PEMDAS.
7
6
5
4
9 – 24 ÷ 8 × 2 + 3
= 9 – 3 × 2 + 3 (Notice that we did division before multiplication because we should go from left to right.)
= 9 – 6 + 3
= 3 + 3 (Notice that we did subtraction before addition because we should go from left to right.)
= 6
3
#### Simplify [(32 ÷ 4) + 3] × 2 using PEMDAS.
20
18
22
10
[(32 ÷ 4) + 3] × 2
= [8 + 3] × 2
= 11 × 2
= 22
4
#### Simplify \$(3 × 5² ÷ 5)\$ – \$(16 — 10)\$ using PEMDAS.
15
9
3
\$(3 × 5² ÷ 5)\$ – \$(16 – 10)\$
= \$(3 × 25 ÷ 5)\$ – \$(16 – 10)\$
= \$(75 ÷ 5)\$ – \$(6)\$
= 15 – 6
= 9
### Frequently Asked Questions On Order Of Operations
What is the order of operations in math?
The order of operations are the rules that tell us the sequence in which we should solve an expression with multiple operations.
The order is PEMDAS: Parentheses, Exponents, Multiplication, and Division (from left to right), Addition and Subtraction (from left to right).
Is there a trick we can use to remember the order of operations?
Yes. You can use the phrase “Please Excuse My Dear Aunt Sally” to remember PEMDAS.
Can we perform subtraction before addition?
Yes, addition and subtraction are at the same level according to the PEMDAS rule. So, without brackets, we do the math from left to right if we are only dealing with addition and subtraction. For example, 9 – 6 + 3 = 3 + 3 = 6.
Can we perform division before multiplication?
Yes, multiplication and division are at the same level according to the PEMDAS rule. So, without brackets, we do the math from left to right if we are only dealing with multiplication and division. For example, 24 ÷ 8 × 2 = 3 × 2 = 6.
Are PEMDAS and BODMAS the same?
Yes. Both PEMDAS and BODMAS are acronyms for remembering the order of operations. They are different names for the same rule. What they call PEMDAS in the US is called BODMAS in the UK, Australia, India and various other countries.
• Division
• Multiplication
• Subtract
## Order of Operations — Math Fun Worksheets
### ORDER OF OPERATIONS WORKSHEETS
These Order of Operations worksheets are the best for children who need to solve mathematical expressions which involves more than one fundamental arithmetic operation. You can also call it PEDMAS worksheets. Thus order of operations are just a set of rules that tell you the order in which math operation (Addition/ Subtraction/ Multiplication/ Division) should be done.
##### Let us learn to solve arithmetic expressions
Let us consider the expression 10- 2 x 3.
If we subtract first 10 — 2 x 3 = 8 x 3 = 24
If we multiply first 10- 2 x 3 = 10- 6 = 4
We are getting two different answers. Hence, to avoid confusions and always arrive the correct answer, we follow a definite order to evaluate an expression.
The rules to follow when you evaluate an arithmetic expression.
1. Do operations in Parentheses first
2. Exponents
3. Multiplication/ Division (Go from left to right, do whichever operation comes first)
4. Addition/ Subtraction (Go from left to right, do whichever operation comes first)
Form an acronym, PEDMAS rule of Operations.
A few arithmetic expressions have been evaluated step- by- step using PEDMAS for easy understanding. Have a look before you download our free worksheets for practice.
#### Multiple choice questions
POPULAR TAGS : Order of operations worksheet, Order of operations practice worksheets, Order of operations problems worksheets
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Admission to the 5th profile class with in-depth study of mathematics
atiki) in the 2023-2024 academic year
(Minutes dated 05/30/2023)
Applications for admission to grade 5 with in-depth study of certain subjects (mathematics) for the 2023-2024 academic year are accepted from June 1 to June 5 inclusive in electronic form (to the school’s email address) or through the class teacher. The application form is attached.
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What is the properties of triangle ?
Geometric figures are sets of points and lines that form shapes. Knowing how to identify triangles is important to understanding the basic properties of triangles. Introduce your child to the concept of triangles with these great introductory worksheets for grade 4 to 7.
Does your child know how to classify triangles? In this article, you will know about the properties of triangles. Geometry worksheet will help to recognize angles and sides of triangles.
Properties of triangle
A triangle is a simple closed curve made of three line segments. Triangle has three vertices,three sides and three angles.
How to classify triangles
We can classify triangles according to their Sides and angles.
Types of triangles based on their sides
1.Equilateral triangle: If the triangle has three equal sides, and therefore three equal angles, then the triangle is called Equilateral triangle.
2.Isosceles triangle: If the triangle has two equal sides, and one unequal, it is isosceles triangle.
3. Scalene triangle: If a triangle has sides of three different lengths, and therefore three different angles, we call it a scalene triangle.
Types of triangles based on their angles
1. Acute-angled triangle: A triangle having all acute angles (less than 90°) in its interior is Acute-angled triangle.
2. Right-angled triangle: A triangle that has a right angle in its interior is Right-angled triangle.
3. Obtuse-angled triangle: A triangle having an obtuse angle
(greater than 90° but less than 180°) in its interior Obtuse-angled triangle.
What is the median of a triangle
A median of a triangle is a line from a vertex of the triangle to the midpoint of the side opposite that vertex. Every triangle have 3 medians. These three medians meet at one point - centered of the triangle.
What is the altitude of a triangle
In a triangle, a segment drawn through a vertex Perpendicular to the opposite side is called the altitude or hight of the triangle. The three altitudes of any triangle are concurrent.
Activities for triangle concept:
Activities help students become familiar with the basic principals and concepts of Geometry.
1. Triangle with strings:
First students have to cut two pieces of string in identical lengths. They then use a protractor to connect the strings to the paper to create a triangle. Once the triangle is constructed, students have to measure each side of the triangle and define its properties.
2.Cardboard triangles:
Make a several cut out of triangles with cardboard in different sizes. Then measure each triangle by their sides and their angles.
Try to classify triangles as you can.
Here you will get many worksheets for practice concept of triangle and its properties. These math worksheets help make learning engaging for your child! Browse through and download our Maths worksheets to get better Education.
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## Sequences-Patterns
A sequence is a set or ordered list of objects or events. A sequence contains numbers or terms which are said
to be the length of the sequence. A set of odd numbers is an example of a sequence.
There are two types of sequences: finite sequences and infinite sequences. A finite sequence is a set of finite numbers.
Example: S = {1, 2, 3……., 10}. An infinite sequence is a sequence in which infinite numbers are presented. Example: a set of even numbers, S = {2, 4, 6, 8……….}.
A pattern may be defined and extended by recursively determining the "next" term. Patterns depend upon the initial condition where the sequence starts and an equation that explains how a term in a sequence can be found from the preceding term.
Example
A sequence between a set of numbers to represent a linear function:
Let us take an example of the given figure in which diagram 1 consists of 3 squares, diagram 2 consists of 5 squares, diagram 3 consists of 7 squares and so on.
If the pattern continues, how many squares will be there in diagram 50 or diagram 100?
Solution
To solve this, we first analyze the number squares in diagram 1 which is equal to 3; then we find the increment of the
squares in the next diagram. In diagram 2, the number of squares increases by 2. We see that as the diagram number
increases, 2 squares in each diagram also increase. If we solve for this then:
No. of diagram 1 2 3 4 5 No. of squares 3 5 7 9 11 Mathematical expression 3+2(0) 3+2(1) 3+2(2) 3+2(3) 3+2(4)
Let us calculate accordingly to find the number of squares in diagram 50 or in diagram 100. As there is an increment of 2
squares with each diagram and the initial diagram has 3 squares, we construct a linear equation to solve this.
We know that the basic form of a linear equation is f (x) = mx + c.
If we compare this expression from the table above, then we find that here
c =3, m = 2 and x= 0, 1, 2, 3, 4…….
The value of x depends upon the diagram number in such a way that if the diagram number is n then the value of x is equal to (n-1).
• uf (n-1) = 3 + 2(n-1) uwhereu n = diagram number.u
Hence, the number of squares in diagram 50 = 3 + 2(50-1)
= 3 + 98
= 101 squares
And the number of squares in diagram 100 = 3 + 2(100-1).
= 3 + 198
= 201 squares
#### General Formula for Arithmetic Sequences between a Set of Numbers
As discussed in the example above, if we take the first term as a1 and the common difference as d, then the second
term is a1 + d, the third term is a1 + 2d and so on……..and the nth term is a1 + (n-1) d.
an = a1 + d (n-1)
where an = the last term of the sequence
a1 = the first term of the sequence
n = the number of terms
d = the common difference between the two successive terms
Here's the simpler formula:
d = a2 - a1
= a3 - a2
General form:
D = an – an-1
Hence, the arithmetic series is an example of a linear function in which the dependent quantity is an and the independent
quantity is n.
Example
Find the 10th term of the arithmetic sequence -10, -5, 0, 5 ….
Solution
We know that
an = a1 + d (n-1)
d = -5-(-10) = 5
n =10
a1 = -10
According to formula, a10 = -10 + 5(10-1).
a10 = -10 + 5*9
a10 = -10 + 45
a10 = 35
#### Try these questions
1. What is the next term in the given pattern?
5, 11, 17, 23, ___
(a) 25
(b) 29
(c) 34
(d) 28
a = 5 d = 6
a5 = 23 + 6 = 29
2. The first five terms of a linear sequence where a = -6 and d = 4 are
(a) -6, -2, 2, 6, 10
(b) -2, 2, 6, 10, 14
(c) -6, 0, 6, 12, 18
(d) 4, 10, 16, 22, 28
a1 = -6
a2 = -6 + 4 = -2
a3 = -2 + 4 = 2
a4 = 2 + 4 = 6
a5 = 6 + 4 = 10
3. What is the 11th term of the given sequence?
2, 5, 8, 11……
(a) 32
(b) 24
(c) 11
(d) 39
a = 2 d = 3
an = a + (n-1) d
a11 = 32
4. Find the linear sequence where an = 9n – 1.
(a) 0, 8, 17…….
(b) 9, 8, 7……..
(c) 8, 17, 26…….
(d) 8, 19, 28………
an = 9n – 1
a1 = 9 – 1 = 8
a2 = 9*2 – 1 = 17
a3 = 9*3 – 1 = 26
a4 = 9*4 – 1 = 35
Hence the sequence is 8, 17, 26, 35…….
5. Discuss in detail the difference between a finite and an infinite sequence with the help of an example.
When the number of objects in a sequence are finite, i.e. when they are countable, then it is called a Finite Sequence. Let us consider a sequence: 2, 3, 4, 5, 6, 7. In the given series, we know that the count of the number of terms is 6. Hence, it is a finite sequence. Let us take another sequence: 5, 8, 11, …, 65. In this example, we can see that THE last term is present; hence it is also a finite sequence. So, if the last term of a sequence is known or we count the number of terms in that sequence, then it is a finite sequence.
When the number of objects present in a sequence are infinite, then the sequence is called an Infinite Sequence.
Let us take an example of such a kind of a sequence: 4, 8, 12… In this example, we cannot count the number of terms as we do not know the last term of the series; hence, it is an infinite sequence. An infinite sequence is very useful for studying vector space, complex analysis, etc. |
Intercepts
There are some key features of graphs that we can use to help us create graphs more quickly and easily. The intercepts are absolutely key points on a graph, and meaningful algebraically, as well. A function will never have more than one y – intercept, but can have many x – intercepts. However, linear equations have only one x – intercept.
Here are some key pieces of information about intercepts.
1. An intercept is where the graph crosses an axis.
1. There are two axes, the x, which is horizontal and the y, which is vertical.
2. For all x – intercepts, the y – coordinate is 0.
1. To find an x – intercept, replace y with zero and solve for x.
3. For all y – intercepts, the x – coordinate is 0.
1. To find a y – intercept, replace x with zero and solve for y.
Because it is easy to do math with zeros, it is a good idea to find the intercepts when using a t-chart!
Universal: Each type of function and equation has its own patterns and short cuts. In all of them, the values of x and or y being zero are of particular importance. The intercepts play key roles and are almost always one of the first pieces of information to be found. With linear equations, we frequently find the y – intercept because it is easy to do so, frequently does not require any calculation at all. Our focus here is linear equations, but the process is the same for all equations. Let’s do a few examples.
Find the intercepts of 4x – 3y = 24.
x – intercept: This will be a coordinate (#, 0). So, y will be zero. Replace y with zero in the equation, then solve for x.
Note: This is not the x – intercept, but the x – coordinate of the x – intercept. The intercept is (6, 0).
y – intercept: This will be a coordinate (0, #). So, x will be zero. Just replace x with zero and solve for y.
Note: This is not the x – intercept, but the x – coordinate of the x – intercept. The intercept is (6, 0).
y – intercept: This will be a coordinate (0, #). So, x will be zero. Just replace x with zero and solve for y.
Note: This is not the y – intercept, but the y – coordinate of the y – intercept. The intercept is (0, -8).
If you know the two intercepts, you can sketch a graph of the line! Here we know (6, 0) and (0, -8) are the intercepts. If we plot those two points, we can connect the dots and have a graph. We might wish to check a random point on the line, plug in the x and y to see if it is a solution. But, if after checking we find a third point is a solution, we can be sure our graph is correct.
This is Graphing Method #2 of graphing a line, Finding the Intercepts.
As has proven to be the case with linear equations, there are a few exceptions. There are three times when this method does not provide quite enough information. You can probably guess two of them, the horizontal and vertical line. Here’s how to handle those equations.
Graphing a Vertical Line
If you’re asked to graph x = #, recognize it will be a vertical line. Every coordinate on the line will have an x component that is the #. So, for x = 14, all coordinates will be (14, some number for y). Just place two points with an x – coordinate of 14, and connect the dots. Extend your line to each end of the coordinate plane.
There will NOT be a y – intercept for a vertical line. However, if the line is x = 0, it will be the y – axis.
Graphing a Horizontal Line
If you’re asked to graph y = #, recognize that it will be horizontal. Every coordinate on that line will have a y component that is the #. So, if you had y = 81, every coordinate would be (some number for x, 81). Just place two points on the with a y – coordinate of 81, and connect the dots. Extend the line to each end of the coordinate plane.
There will NOT be an x – intercept for a horizontal line. However, if the line is y = 0, it will be the x – axis.
The Origin
The third time finding the intercepts of a linear equation fails to provide enough information to make a graph is if the intercepts pass through the origin. In this case, you’ll need to either use a t-chart, count your slope (you’ll see that in the next section), or just find another point (explained soon).
Here’s an example:
y = 2x
The x – intercept is (0, 0). Here’s the work: (0) = 2xx = 0.
The y – intercept is (0, 0). Here’s the work: y = 2(0) → y = 0.
Even though we found both intercepts, they’re at the same exact spot. We need two points to draw a line (without guessing its direction). In this case, probably the easiest thing to do is just figure out another solution. The equation says that the output (y) is twice as big as the input (x). To make them equal, the x is doubled. So, if x was 3, y would be 6. (3, 6) will be a point on the line. Plot (0, 0), and then (3, 6), and you’re done!
Summary: A graph is a picture of all solutions to an equation. If you know the shape of the equation (linear equations form a line), then you only need a few clues as to where the graph will be. A t-chart is a sure-fire, always works, way to find out information for any equation! The next method, that is almost as reliable, and will translate to almost any type of graph is finding the intercepts. For an x – intercept, y = 0, and for a y – intercept, x = 0.
There are two strange cases with linear equations, the vertical and horizontal lines. The line y = 3, for example, is horizontal because every coordinate on that line will be some number for x and y = 3, like (2, 3), (0, 3) and (-11, 3). It will have a y – intercept, but not an x – intercept.
A vertical line will be x equals a number, like x = 3. This will have an x – intercept, but not a y – intercept. Every coordinate on this line will have x = 3, and y can be any number. |
Table of Contents
## What is indicated probability formula?
P(A or B) = P(A) + P(B) – P(A and B)
How do you find probability probabilities?
How to calculate probability
1. Determine a single event with a single outcome.
2. Identify the total number of outcomes that can occur.
3. Divide the number of events by the number of possible outcomes.
How do you find the indicated probability for the normal distribution variable?
The probability that a standard normal random variables lies between two values is also easy to find. The P(a < Z < b) = P(Z < b) – P(Z < a). For example, suppose we want to know the probability that a z-score will be greater than -1.40 and less than -1.20.
### What is standard normal probability distribution?
The standard normal distribution is a normal distribution with a mean of zero and standard deviation of 1. The standard normal distribution is centered at zero and the degree to which a given measurement deviates from the mean is given by the standard deviation.
How do you find the normal distribution?
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All you have to do to solve the formula is:
1. Subtract the mean from X.
2. Divide by the standard deviation.
How do I find the standard deviation?
To calculate the standard deviation of those numbers:
1. Work out the Mean (the simple average of the numbers)
2. Then for each number: subtract the Mean and square the result.
3. Then work out the mean of those squared differences.
4. Take the square root of that and we are done!
## How do you find the mean and standard deviation from a frequency table?
Use this step-by-step approach to find the standard deviation for a discrete variable.
1. Calculate the mean.
2. Subtract the mean from each observation.
3. Square each of the resulting observations.
4. Add these squared results together.
5. Divide this total by the number of observations (variance, S2).
How do you find the standard deviation in a frequency distribution table?
The mean is the sum of the product of the midpoints and frequencies divided by the total of frequencies. Simplify the right side of μ=1368 μ = 136 8 . The equation for the standard deviation is S2=∑f⋅M2−n(μ)2n−1 S 2 = ∑ f ⋅ M 2 – n ( μ ) 2 n – 1 .
How do you solve for grouped data in statistics?
To calculate the mean of grouped data, the first step is to determine the midpoint (also called a class mark) of each interval, or class. These midpoints must then be multiplied by the frequencies of the corresponding classes. The sum of the products divided by the total number of values will be the value of the mean.
### How do you find the mean and standard deviation of grouped data in Excel?
But first, let us have some sample data to work on:
1. Calculate the mean (average)
2. For each number, subtract the mean and square the result.
3. Add up squared differences.
4. Divide the total squared differences by the count of values.
5. Take the square root.
6. Excel STDEV function.
7. Excel STDEV.
8. Excel STDEVA function. |
# Associative Property of Multiplication – Definition With Examples
Table of Contents
Understanding mathematical concepts is crucial for children’s development, and one fundamental concept they encounter is the Associative Property of Multiplication. This property, which forms the foundation of multiplication, allows children to explore how numbers can be grouped and how they relate to addition. In this article, brought to you by Brighterly, we will delve into the definition of the Associative Property of Multiplication and provide engaging examples to enhance comprehension.
## What is the Associative Property of Multiplication?
The Associative Property of Multiplication states that when multiplying three or more numbers, the grouping of the numbers does not affect the result. In simpler terms, it means that changing the order in which we multiply a set of numbers doesn’t change the final outcome.
## Associative Property of Multiplication Formula
The formula for the Associative Property of Multiplication can be expressed as:
(a * b) * c = a * (b * c)
Here, a, b, and c represent any real numbers that we want to multiply together. This formula demonstrates that regardless of how we group the numbers, the product will remain the same.
## Associative Property of Multiplication and Addition
It’s worth noting that the Associative Property of Multiplication is related to the Associative Property of Addition. Both properties involve grouping and changing the order of operations while preserving the final result. These properties are essential in simplifying complex mathematical expressions.
Now, let’s dive into some examples to illustrate the concept more vividly.
## Examples on Associative Property of Multiplication
Example 1: Consider the numbers 2, 3, and 4. According to the Associative Property of Multiplication, we can group them in different ways without altering the result:
(2 * 3) * 4 = 6 * 4 = 24
2 * (3 * 4) = 2 * 12 = 24
In both cases, the final product is 24, demonstrating the Associative Property of Multiplication.
Example 2: Let’s take a more complex example: 5, 6, 7, and 8. Applying the Associative Property of Multiplication, we can rearrange the grouping as follows:
((5 * 6) * 7) * 8 = (30 * 7) * 8 = 210 * 8 = 1680
5 * ((6 * 7) * 8) = 5 * (42 * 8) = 5 * 336 = 1680
Again, the result remains the same regardless of the grouping.
## Practice Questions on Associative Property of Multiplication
1. Apply the Associative Property of Multiplication to simplify: (3 * 4) * 2 =
2. Demonstrate the Associative Property of Multiplication using the numbers 9, 10, and 11.
3. True or False: The Associative Property of Multiplication can be applied to any set of numbers.
4. Rewrite the expression (2 * 3) * (4 * 5) using the Associative Property of Multiplication.
## Conclusion
In conclusion, the Associative Property of Multiplication is a fundamental concept in mathematics that allows children to understand how multiplication can be grouped and how it relates to addition. By grasping this property, children gain a deeper understanding of how numbers interact and how to simplify complex multiplication problems.
Throughout this article, we have explored the definition of the Associative Property of Multiplication and provided numerous examples to reinforce comprehension. By practicing with these examples and understanding the underlying concept, children can strengthen their mathematical skills and problem-solving abilities.
At Brighterly, we believe in making math enjoyable and accessible for children. By introducing concepts like the Associative Property of Multiplication in a creative and engaging manner, we empower children to develop a strong foundation in mathematics. Encouraging children to explore the properties and patterns of numbers will not only enhance their mathematical abilities but also foster critical thinking and analytical skills that will benefit them throughout their lives.
## Frequently Asked Questions on the Associative Property of Multiplication
### Why is the Associative Property of Multiplication important?
The Associative Property of Multiplication is important because it simplifies calculations and allows us to rearrange the grouping of numbers without affecting the final result. This property is particularly useful when dealing with more complex mathematical operations. By leveraging the Associative Property, we can break down large multiplication problems into smaller, more manageable parts, making calculations faster and more efficient.
### Can the Associative Property of Multiplication be applied to any set of numbers?
Yes, the Associative Property of Multiplication holds true for any set of real numbers. Whether we’re multiplying whole numbers, fractions, decimals, or even negative numbers, we can apply the Associative Property to rearrange the grouping without altering the final product. This property is a fundamental rule in mathematics that applies universally, allowing for consistent and reliable calculations across various numerical contexts.
### How does the Associative Property of Multiplication relate to the real world?
The Associative Property of Multiplication finds applications in various fields, demonstrating its practical significance in the real world. For instance, in physics, when calculating the force exerted by multiple interacting objects, the ability to regroup the multiplication of factors simplifies the calculations and provides a clearer understanding of the forces at play. In economics, when analyzing the impact of different factors on a product’s price or production costs, the Associative Property allows for efficient evaluation of complex scenarios. In computer science and programming, this property plays a vital role in optimizing algorithms and improving computational efficiency.
Sources:
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• Grade 1
• Grade 2
• Grade 3
• Grade 4
• Grade 5
• Grade 6
• Grade 7
• Grade 8 |
Identities of Complex Numbers
A complex number is a number that is made up of both real and imaginary numbers. Here we talk about some properties of complex numbers. To familiarize more with them read this post.
A complex number is written as $$a+ib$$ and usually represented by $$z$$. Where $$a$$ signifies a real number and $$ib$$ represents an imaginary number. In addition, $$a,b$$ are real values, and $$i^2 = -1$$.
As a result, a complex number is a straightforward representation of the addition of two integers, namely a real and an imaginary number. One side is entirely genuine, while the other is entirely imagined.
A step-by-step guide to identities of complex numbers
The following are some of the properties of complex numbers:
• The sum of two conjugate complex numbers will result in a real number.
• The multiplying of two conjugate complex numbers will produce a real number as well as a complex number.
• If $$x$$ and $$y$$ are real numbers and $$x+yi =0$$, then $$x =0$$ and $$y =0$$ are the same value.
• When two conjugate complex numbers are added together, the result is a real number.
• A real number can be obtained by multiplying two conjugate complex numbers.
• If $$p$$, $$q$$, $$r$$, and $$s$$ are real numbers, then $$p+qi = r+si$$, $$p = r$$, and $$q=s$$.
• The “basic law” of addition and multiplication applies to complex numbers: $$\color{blue}{z_{1 }+ z_{2 }= z_{2 }+ z_{1 } }$$ , $$\color{blue}{z_{1 }. z_{2 }= z_{2 }. z_{1 } }$$
• The complex number follows the “associative law” of addition and multiplication, which is a mathematical rule: $$\color{blue}{(z_{1 }+ z_{2 })+ z_{3 }= z_{1 }+ (z_{2 }+z_{3 })}$$ ,$$\color{blue}{(z_{1 }. z_{2 }).z_{3 }=z_{1 } (z_{2 }. z_{3 } )}$$
• The “distributive law” applies to complex numbers: $$\color{blue}{z_{1 }( z_{2 }+z_{3 })= z_{1 }. z_{2 }+z_{1 }z_{3 } }$$
• In other words, if the sum of two complex numbers is real, and the product of two complex numbers is also genuine, then these complex numbers are conjugated with one another.
• For any two complex numbers, say $$z_{1 }$$ and $$z_{2 }$$, then $$|z_{1 }+z_{2 }|≤|z_{1 }|+|z_{2 }|$$
• When two complex numbers are multiplied by their conjugate value, the output should be a complex number with a positive value.
Algebraic identities of complex numbers
All algebraic identities apply equally to complex numbers. The addition and subtraction of complex numbers with the exponents of $$2$$ or $$3$$ can be easily solved using algebraic identities of complex numbers.
• $$\color{blue}{(z_{1 }+ z_{2 })^2= (z_{1 })^2+ (z_{2 })^2+2 z_{1 }× z_{2 } }$$
• $$\color{blue}{(z_{1 }- z_{2 })^2= (z_{1 })^2+ (z_{2 })^2-2 z_{1 }× z_{2 } }$$
• $$\color{blue}{(z_{1 })^2- (z_{2 })^2= (z_{1 }+ z_{2 })( z_{1 }- z_{2 }) }$$
• $$\color{blue}{(z_{1 }+ z_{2 })^3= (z_{1 })^3+ 3(z_{1 })^2 z_{2 } +3 (z_{2 })^2 z_{1 }+ (z_{2 })^3 }$$
• $$\color{blue}{(z_{1 }- z_{2 })^3= (z_{1 })^3- 3(z_{1 })^2 z_{2 } +3 (z_{2 })^2 z_{1 }- (z_{2 })^3 }$$
Identities of Complex Numbers – Example 1:
Find the sum of the complex numbers. $$z_{1 }=-3+i$$ and $$z_{2 }=4-3i$$
$$z_{1 }$$ $$+$$ $$z_{2 }$$ $$=(-3+i)+(4-3i)=(-3+4)+(i-3i)=1-2i$$
Identities of Complex Numbers – Example 2:
Solve the complex numbers $$(2+i)^2$$.
To solve complex numbers use this formula: $$\color{blue}{(z_{1 }+ z_{2 })^2= (z_{1 })^2+ (z_{2 })^2+2 z_{1 }× z_{2 } }$$
$$(2+i)^2$$ $$=$$ $$(2)^2$$$$+$$$$(i)^2$$$$+(2×2×i)$$$$=4+i^2+4i$$
Then: $$i^2=-1$$ → $$4+i^2=4-1=3$$
Now: $$4+i^2+4i =3+4i$$
Identities of Complex Numbers – Example 3:
Solve the complex numbers $$(3-i)^3$$.
First, use this formula: $$\color{blue}{(z_{1 }- z_{2 })^3= (z_{1 })^3- 3(z_{1 })^2 z_{2 } +3 (z_{2 })^2 z_{1 }- (z_{2 })^3 }$$
$$(3-i)^3$$ $$=(3)^3-3(3)^2(i)+3(i)^2(3)-(i)^3$$ $$=27-27i+9i^2-i^3$$
Then: $$i^2=-1$$ → $$9i^2=-9$$
$$=27-27i-9-i^3$$ $$=18-27i-i^3$$
$$i^3=-i$$ → $$18-27i-i^3= 18-27i-(-i)=$$ $$18-27i+i$$
Now: $$18-27i+i =18-26i$$
Exercises for Identities of Complex Numbers
Simplify.
1. $$\color{blue}{(4+5i)^2}$$
2. $$\color{blue}{(12+5i)+(3+i^2+6i)}$$
3. $$\color{blue}{(20+7i)-(45i+12)}$$
4. $$\color{blue}{(5-4i)^2(3+3i)}$$
5. $$\color{blue}{(i^2-5i)^3}$$
6. $$\color{blue}{(6-i)^2-(10+i)^2}$$
1. $$\color{blue}{-9+40i}$$
2. $$\color{blue}{14+11i}$$
3. $$\color{blue}{8-38i}$$
4. $$\color{blue}{147-93i}$$
5. $$\color{blue}{74+110i}$$
6. $$\color{blue}{-64-32i}$$
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Unit 7 Section 1 : Shapes
# Unit 7 Section 1 : Shapes
You should be familiar with the common 2-D shapes, but to recap, we give the names and definitions below.
## Example 1
What could each one of the following shapes be if it has 4 sides and:
(a) opposite sides equal and parallel, It could be a parallelogram, rhombus, rectangle or square. all sides equal, It could be a rhombus or square. two adjacent angles are right angles? It could be a trapezium, rectangle or square.
## Example 2
For the grid opposite, name all shapes that are:
(a)
congruent,
Congruent to A means the same size and shape as A. The shapes congruent to A are G, L and K.
(b)
similar to shape A.
Similar to A means the same shape as A but not necessarily the same size as A. The shapes similar to A are C, F, G, J, K and L.
## Example 3
Using 20 m of fencing, design four different rectangular enclosures. For each one, find its area. Which shape gives the maximum area?
Possible shapes could be:
area = 2 × 8 = 16m² area = 3 × 7 = 21m² area = 4 × 6 = 24m² area = 5 × 5 = 25m²
The square (5 m × 5 m) gives the maximum area.
## Exercises
Question 1
What could each one of the following shapes be if it has 4 sides and:
(a)
all angles are right angles,
(b)
only one pair of opposite sides are parallel, but not equal,
(c)
diagonals intersect at right angles?
Question 2
Which of the shapes in the diagram below are:
(a) (b) congruent, similar
to shape A ?
List the numbers separated by commas: e.g. A, B, C, D
(a)
(b)
Question 3
Using 40 cm of wire, design different rectangles. For each one, find its area.
What shape gives the maximum area?
The shape that gives the maximum area is a of side cm.
Question 4
These two congruent triangles make a parallelogram.
(a)
On the grid below, draw another congruent triangle to make a rectangle.
(b)
On a copy of the grid below, draw another congruent triangle to make a bigger triangle.
(c)
On a copy of the grid below, draw another congruent triangle to make a different bigger triangle.
Question 5
Mike has a triangle grid. He shades in 2 triangles to make a shape with 4 sides.
(a)
Shade in 2 triangles on a grid below to make a different shape with 4 sides.
(b)
On another grid, shade in 2 triangles to make another different shape with 4 sides.
(c)
On another grid, shade in 4 small triangles to make a bigger triangle.
(d)
On the grid below, shade in more than 4 small triangles to make a bigger triangle.
Question 6
(a)
Scott has 9 small square tiles. On the following grid, show how Scott can make a square in the same way with 9 small square tiles.
(b)
On another grid, show how to make a square with more than 9 of these small square tiles.
How many tiles are there in your square?
(c)
Huw wants to make some more squares with the tiles. Write down 3 other numbers of tiles that he can use to make squares.
Question 7
Helen has these eight rods.
She can use 5 of her rods to make a rectangle.
This combination is: 5, 5, 8, 6 + 2
(a)
On a copy of the grid above, show how to make a different rectangle with a different shape with 5 of Helen's rods.
, , ,
(b)
On a larger grid, 13 squares by 10 squares, show how to make a rectangle with 6 of Helen's rods.
, , ,
(c)
On another large grid, show how to make a square with all 8 of Helen's rods.
, , , |
<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
# Conservation of Momentum in Two Dimensions
## Apply component vectors to solve two dimensional collision problems
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Conservation of Momentum in Two Dimensions
In a game of billiards, it is important to be able to visualize collisions in two dimensions – the best players not only know where the target ball is going but also where the cue ball will end up.
### Conservation of Momentum in Two Dimensions
Conservation of momentum in all closed systems is valid, regardless of the directions of the objects before and after they collide. Most objects are not confined to a single line, like trains on a rail. Rather, many objects, like billiard balls or cars, can move in two dimensions. Conservation of momentum for these objects can also be calculated; momentum is a vector and collisions of objects in two dimensions can be represented by axial vector components. To review axial components, revisit Vectors: Resolving Vectors into Axial Components and Vectors: Vector Addition.
Example Problem: A 2.0 kg ball, , is moving with a velocity of 5.00 m/s due west. It collides with a stationary ball, , also with a mass of 2.0 kg. After the collision, ball moves off at 30° south of west while ball moves off at 60° north of west. Find the velocities of both balls after the collision.
Solution: Since ball is stationary before the collision, then the total momentum before the collision is equal to momentum of ball . The momentum of ball before collision is shown in red below, and can be calculated to be
Since momentum is conserved in this collision, the sum of the momenta of balls and after collsion must be 10.0 kg m/s west.
To find the final velocities of the two balls, we divide the momentum of each by its mass. Therefore, and .
Example Problem: A 1325 kg car moving north at 27.0 m/s collides with a 2165 kg car moving east at 17.0 m/s. The two cars stick together after the collision. What is the speed and direction of the two cars after the collision?
Solution:
Example Problem: A 6.00 kg ball, , moving at velocity 3.00 m/s due east collides with a 6.00 kg ball, , at rest. After the collision, moves off at 40.0° N of E and ball moves off at 50.0° S of E.
1. What is the momentum of after the collision?
2. What is the momentum of after the collision?
3. What are the velocities of the two balls after the collision?
Solution:
This is a right triangle in which the initial momentum is the length of the hypotenuse and the two momenta after the collision are the legs of the triangle.
#### Summary
• The conservation of momentum law holds for all closed systems regardless of the directions of the objects before and after they collide.
• Momentum is a vector; collisions in two dimensions can be represented by axial vector components.
#### Practice
Questions
This video shows circus performers using conservation of momentum. Use this resource to answer the questions that follow.
1. Why do the fliers scrunch up in the air while spinning and twisting?
2. What happens to the rate at which they spin when they change shape in the air?
#### Review
Questions
1. Billiard ball , mass 0.17 kg, moving due east with a velocity of 4.0 m/s, strikes stationary billiard ball , also mass of 0.17 kg. After the collision, ball moves off at an angle of 30° north of east with a velocity of 3.5 m/s, and ball moves off at an angle of 60° south of east. What is the speed of ball ?
2. A bomb, originally sitting at rest, explodes and during the explosion breaks into four pieces of exactly 0.25 kg each. One piece flies due south at 10 m/s while another pieces flies due north at 10 m/s.
1. What do we know about the directions of the other two pieces and how do we know it?
2. What do we know about the speeds of the other two pieces and how do we know it?
3. In a head-on collision between protons in a particle accelerator, three resultant particles were observed. All three of the resultant particles were moving to the right from the point of collision. The physicists conducting the experiment concluded there was at least one unseen particle moving to the left after the collision. Why did they conclude this? |
Converting measurements from one unit to another is a common task in various fields, including construction, engineering, and everyday life. One such conversion that often arises is converting centimeters to feet. In this article, we will explore the process of converting 156 centimeters to feet, providing valuable insights and examples along the way.
## Understanding the Basics: Centimeters and Feet
Before diving into the conversion process, let’s first understand the units involved.
### Centimeters (cm)
Centimeters are a unit of length in the metric system. They are commonly used to measure smaller distances, such as the height of a person or the length of an object. One centimeter is equal to one-hundredth of a meter.
### Feet (ft)
Feet, on the other hand, are a unit of length in the imperial system. They are widely used in the United States and a few other countries. One foot is equal to 12 inches or approximately 30.48 centimeters.
## The Conversion Process: 156 cm to Feet
Now that we have a clear understanding of the units involved, let’s proceed with the conversion of 156 centimeters to feet.
To convert centimeters to feet, we need to divide the number of centimeters by the conversion factor, which is 30.48 (the number of centimeters in a foot). Let’s calculate:
156 cm ÷ 30.48 = 5.118 ft
Therefore, 156 centimeters is approximately equal to 5.118 feet.
## Real-World Examples
Understanding the conversion process is essential, but let’s explore some real-world examples to solidify our understanding.
### Example 1: Height Conversion
Imagine you are planning a trip to the United States, and you want to know your height in feet. If you are 156 centimeters tall, you can use the conversion process we discussed earlier to find your height in feet:
156 cm ÷ 30.48 = 5.118 ft
Therefore, your height is approximately 5.118 feet.
### Example 2: Construction Project
Suppose you are working on a construction project that requires precise measurements. One of the dimensions you need to convert is 156 centimeters. By using the conversion process, you can easily convert it to feet:
156 cm ÷ 30.48 = 5.118 ft
Therefore, the dimension is approximately 5.118 feet.
Here are some common questions that often arise when converting centimeters to feet:
### Q1: How accurate is the conversion from centimeters to feet?
A1: The conversion from centimeters to feet is accurate up to the decimal places provided. However, it’s important to note that rounding may be necessary depending on the level of precision required for a particular application.
### Q2: Can I use an online converter to convert centimeters to feet?
A2: Yes, there are numerous online converters available that can quickly and accurately convert centimeters to feet. These converters are especially useful when dealing with large numbers or when precision is crucial.
### Q3: Are there any other common units used to measure length?
A3: Yes, apart from centimeters and feet, there are several other units used to measure length. Some examples include meters, inches, yards, and miles. The choice of unit depends on the specific application and the country or region where the measurement is being made.
### Q4: How can I convert feet to centimeters?
A4: To convert feet to centimeters, you need to multiply the number of feet by the conversion factor, which is 30.48 (the number of centimeters in a foot). For example, if you have 5 feet, the conversion would be:
5 ft × 30.48 = 152.4 cm
### Q5: Can I use the conversion factor for centimeters to feet in reverse?
A5: Yes, the conversion factor can be used in reverse to convert feet to centimeters. By dividing the number of feet by 0.0328084 (approximately 30.48), you can convert feet to centimeters.
## Summary
Converting centimeters to feet is a straightforward process that involves dividing the number of centimeters by the conversion factor of 30.48. In this article, we explored the conversion of 156 centimeters to feet, providing real-world examples and answering common questions along the way. Whether you are planning a trip, working on a construction project, or simply curious about your height in feet, understanding this conversion process can be valuable in various situations.
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## Learning Opportunities
This puzzle can be solved using the following concepts. Practice using these concepts and improve your skills.
## Goal
You are Catherine, you know that Alice and Bob are chatting secretly and you want to be the woman in the middle; ie you want to intercept Alice and Bob’s messages, to read them and to send them messages that you cipher in place of theirs. If Alice sent Bob a message, Bob will never receive Alice’s message but yours instead.
You are lucky: you caught a ciphered text and its clear version and you know that they are using the Hill cipher method.
You have to find the Hill matrix in order to decipher a text from Alice or Bob and to cipher the text you want to send instead.
The texts are using the 45 symbols of the alphanumeric encoding of the QR codes:
0123456789ABCDEFGHIJKLMNOPQRSTUVWXYZ \$%*+-./: (yes, it’s a space between the Z and the \$)
How to:
The Hill cipher uses matrix multiplication (here, with a modulus of 45).
For example, you have to cipher CODINGAME. with A, a 2×2 matrix.
CODINGAME. is 12 24 13 18 23 16 10 22 14 42 (because C’s index is 12).
If A is:
`⎡6 5⎤⎣7 6⎦`
The first pair of ciphered symbols will be:
6×12+5×24 mod 45 = 192 mod 45 = 12 that is C and 7×12+6×24 mod 45 = 228 mod 45 = 3 that is 3.
Thus, CO is ciphered C3, the product A×T mod 45 where T is the column matrix whose elements are 12 and 24.
Finally, the ciphered text is C3XJ%WZMOZ.
If the matrix’s size is 3, you will cipher 3 characters at a time.
For this puzzle, you have to find the cipher matrix (and its size) and how to decipher.
Input
Line 1: The ciphered text
Line 2: The clear text
Line 3: The ciphered text you have to decipher
Line 4: The clear text you have to cipher
Output
Line 1: Line 3 deciphered
Line 2: Line 4 ciphered
Constraints
You can assume that the size of the cipher matrix divides the length of all the messages (but it’s not 1).
* The clear texts of a test and validator pair will be the same but the Hill matrix won’t be the same.
* Test 1/validator 1 and test 2/validator 2’s Hill matrices are the same.
* Test 2/validator 2 and test 3/validator 3’s clear texts are the same.
* There are no trailing spaces in the clear texts but there might be some in the ciphered texts (as in the example).
Example
Input
```C3XJ%WZMOZ
CODINGAME.
6-85OXC
HELLO WORLD.```
Output
```BONJOUR.
\$N639O.8.0IS```
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## First Method- regular fractions
Question: 3/5 - 2/3
First, all you do is list the multiples of each denominator... 5,10,15
3,6,9,12,15
As you can see the common denominator is 15,
Now that we know the common denominator we have to multiply
5 multiplied by 3 gives us 15 and 3 multiplied by 5 gives us 15
remember anything you do to the denominator you HAVE to do to the numerator
3 x 3 2 x 5
____ -- ____
5 x 3 3 x 5
so our fractions are now... 9/15 - 10/15
now that we have the denominators the same we can simply subtract the numerators
9-10=1
Answer= 1/15---remember the denominator always stays the same.
Learn Fractions - How to Subtract Fractions
## Second Method- improper fractions
Question: 15/2 - 10/5
First, we find the common denominator
5,10
2,4,6,8,10
As you can see 10 is the common denominator,now its time to multiply
5 multiplied by 2 is 10 and 2 multiplied by 5 is 10
remember anything you do to the denominator you must do to the numerator
15 x 5 10 x 2
_______ --- ________
2 x 5 5 x 2
our fractions are now.... 75/10 - 20/10
After that we simply subtract the numerators
75 - 20 = 55
In this example the answer can be reduced and since this is a improper fraction we have to turn it into a mixed number, so we have to see how many times the denominator goes into the numerator, 10 goes into 55 , 5 times so 5 will be our whole number and the remainder is 5 and the denominator stays the same so it will be 10 so it will look like this .........
Whole #=5
remainder =5
denominator=10
## Third Method- mixed numbers
Question: 1 11/6 - 2 2/3
First, we have to turn each mixed number into an improper fraction
and to do that we have first multiply the whole number and the denominator and then add the numerator so 1 x 6 + 11 = 17/6 and 2 x 3 + 2 = 8/3 ---- remember the denominators always stay the same
Now that we have the improper fractions we must find the common denominators
6,12
3,6
As you can see 6 is the common denominator, 6 x 1 is 6 and 3 x 2 is 6 so...
17 x 1 8 x 2
______ ---- ______
6 x 1 3 x 2
our fractions are: 17/6 - 16/6
now that we have that we can simply subtract the numerators
17-16=1 |
Section8.5Chapter 8 Summary and Review
SubsectionLesson 8.1 Algebraic Fractions
• An algebraic fraction (or rational expression, as they are sometimes called) is a fraction in which both numerator and denominator are polynomials.
• An algebraic fraction is undefined at any values of the variable that make the denominator equal to zero.
• We use the fundamental principle of fractions to reduce or build fractions.
• Fundamental Principle of Fractions.
We can multiply or divide the numerator and denominator of a fraction by the same nonzero factor, and the new fraction will be equivalent to the old one.
\begin{equation*} \blert{\dfrac{a \cdot c}{b \cdot c} = \dfrac{a}{b}~~~\text{if}~~~b,~c \not= 0} \end{equation*}
• To Reduce an Algebraic Fraction.
1. Factor numerator and denominator completely.
2. Divide numerator and denominator by any common factors.
• We can cancel common factors (expressions that are multiplied together), but not common terms (expressions that are added or subtracted).
• A fraction is a negative number if either its numerator or its denominator is negative, but not both.
• Negative of a Binomial.
The opposite of $a-b$ is
\begin{equation*} \blert{-(a-b) = -a+b = b-a} \end{equation*}
SubsectionLesson 8.2 Operations on Algebraic Fractions
• Product of Fractions.
If $~b \not= 0~$ and $~d \not= 0~\text{,}$ then
\begin{equation*} \blert{\dfrac{a}{b} \cdot \dfrac{c}{d} = \dfrac{ac}{bd}} \end{equation*}
• To Multiply Algebraic Fractions.
1. Factor each numerator and denominator completely.
2. If any factor appears in both a numerator and a denominator, divide out that factor.
3. Multiply the remaining factors of the numerator and the remaining factors of the denominator.
4. Reduce the product if necessary.
• Quotient of Fractions.
If $b,~c,~,d \not= 0\text{,}$ then
\begin{equation*} \dfrac{a}{b} \div \dfrac{c}{d} = \dfrac{a}{b} \cdot \dfrac{d}{c} \end{equation*}
• To Divide One Fraction by Another.
1. Take the reciprocal of the second fraction and change the division to multiplication.
2. Follow the rules for multiplication of fractions.
• Fractions with the same denominator are called like fractions.
• Sum or Difference of Like Fractions.
If $c \not= 0\text{,}$ then
\begin{equation*} \blert{\dfrac{a}{c}+\dfrac{b}{c}= \dfrac{a+b}{c}}~~~~~~\text{and}~~~~~~ \blert{\dfrac{a}{c}-\dfrac{b}{c}= \dfrac{a-b}{c}} \end{equation*}
• To Add or Subtract Like Fractions.
1. Add or subtract the numerators.
2. Keep the same denominator.
3. Reduce the sum or difference if necessary.
SubsectionLesson 8.3 Lowest Common Denominator
• To add or subtract fractions with unlike denominators, we must first convert the fractions to equivalent forms with the same denominator.
• The lowest common denominator for two or more algebraic fractions is the simplest algebraic expression that is a multiple of each denominator.
• To Find the LCD.
1. Factor each denominator completely, and arrange the factors in order.
2. For each factor,
1. Which denominator has the most copies of that factor? Circle them. (If there is a tie, either denominator will do.)
2. Include all the circled factors in the LCD.
3. Multiply together the factors of the LCD.
• To Add or Subtract Algebraic Fractions.
1. Find the lowest common denominator (LCD) for the fractions.
2. Build each fraction to an equivalent one with the LCD as denominator.
3. Add or subtract the resulting like fractions: Add or subtract their numerators, and keep the same denominator.
4. Reduce the sum or difference if necessary.
SubsectionLesson 8.4 Equations with Fractions
• To solve an equation that contains algebraic fractions, we first clear the denominators by multiplying both sides of the equation by the LCD of the fractions.
• When clearing fractions from am equation, we must be sure to multiply each term of the equation by the LCD.
• Whenever we multiply an equation by an expression containing the variable, we should check for extraneous solutions.
• Work Formula.
\begin{align*} \blert{\text{work rate} \times \text{time}} \amp \blert{= \text{work completed}}\\ \blert{rt} \amp \blert{= w} \end{align*}
SubsectionReview Questions
Use complete sentences to answer the questions.
1. When reducing an algebraic fraction, we should always before we .
2. The fundamental principle of fractions says that we can cancel , but not .
3. Delbert says that to multiply two algebraic fractions, we just multiply the numerators together and multiply the denominators together. Comment on Delbert's method.
4. To divide by a fraction is the same as to by its.
5. Describe how to add or subtract unlike fractions in three steps.
6. Why do we need to find an LCD when adding unlike fractions?
7. Francine says that to solve an equation containing algebraic fractions, we build each fraction to an equivalent one with the LCD. Comment on Francine's method.
8. When might you expect to encounter extraneous solutions?
9. Which of the operations listed below use an LCD?
2. multiply fractions
3. solve an equation with fractions
4. subtract fractions
5. divide fractions
10. Which of the operations listed above use building factors?
SubsectionReview Problems
ExercisesExercises
1.
1. Evaluate the fraction $~\dfrac{s^2-s}{s^2+3s-10}~~$for $~s=-2$
2. For what value(s) of $s$ is the fraction undefined?
2.
Ed's Diner uses a package of coffee filters every $x+5$ days.
1. What fraction of a package does Ed use every day?
2. What fraction of a package does Ed use in one week?
For Problems 3–10, reduce the fraction if possible.
3.
$\dfrac{a+3}{b+3}$
4.
$\dfrac{5x+7}{5x}$
5.
$\dfrac{10+2y}{2y}$
6.
$\dfrac{3x^2-1}{1-3x^2}$
7.
$\dfrac{v-2}{v^2-4}$
8.
$\dfrac{q^5-q^4}{q^4}$
9.
$\dfrac{-3x}{6x^2+9x}$
10.
$\dfrac{x^2+5x+6}{x^2-4}$
For Problems 11–16,
2. multiply the fractions.
11.
$\dfrac{3}{8},~\dfrac{5}{12}$
12.
$\dfrac{2}{x},~\dfrac{1}{x+2}$
13.
$\dfrac{3x}{2x+2},~\dfrac{x+1}{6x}$
14.
$\dfrac{x+1}{x-1},~\dfrac{1}{x^2-1}$
15.
$2,~\dfrac{1}{x}$
16.
$x,~\dfrac{1}{x+2}$
For Problems 17–22, write the expression as a single fraction in lowest terms.
17.
$\dfrac{4c^2d}{3}\div(6cd^2)$
18.
$\dfrac{u^2-2uv}{uv} \div \dfrac{3u-6v}{2uv}$
19.
$\dfrac{2m^2-m-1}{m+1}-\dfrac{m^2-m}{m+1}$
20.
$\dfrac{3}{2p}+\dfrac{7}{6p^2}$
21.
$\dfrac{5q}{q-3}-\dfrac{7}{q}+3$
22.
$\dfrac{2w}{w^2-4}+\dfrac{4}{w^2+4w+4}$
23.
On Saturday mornings, Olive takes her motorboat 5 miles upstream to the general store for supplies and then returns home. The current in the river is 2 miles per hour. Let $x$ represent Olive's speed in still water, and write algebraic fractions to answer each question.
1. How long does it take Olive to get to the store?
2. How long does the return trip take?
3. How long does the round trip take?
24.
On spring break, Johann and Sebastian both walk from the university to the next town. Johann leaves at noon and Sebastian leaves one hour later, but Sebastian walks 1 mile per hour faster. Let $r$ stand for Johann's walking speed. Write polynomials to answer the following questions.
1. What is Sebastian's walking speed?
2. How far has Sebastian walked at 3 pm?
3. How far has Johann walked at 3 pm?
4. How far apart are Johann and Sebastian at 3 pm?
For Problems 25–28, solve the equation.
25.
$q-\dfrac{16}{q}=6$
26.
$\dfrac{2-x}{5x} =\dfrac{4}{15x}-\dfrac{1}{6}$
27.
$\dfrac{9}{m+2}+\dfrac{2}{m}=2$
28.
$\dfrac{15}{x^2-3x}+\dfrac{4}{x}=\dfrac{5}{x-3}$
For Problems 29–30, solve solve for the indicated variable.
29.
$\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{1}{z}~~~~$for $~z$
30.
$y=\dfrac{2x+3}{1-x}~~~~$for $~x$
For Problems 31–32,
1. Write an equation to model the problem.
2. Solve your equation and answer the question posed in the problem.
31.
On a walking tour, Nora walks uphill 5 miles to an inn where she has lunch. After lunch, she increases her speed by 2 miles per hour and walks for 8 more miles. If she walked for 1 hour longer before lunch than after lunch, what was her speed before lunch?
32.
Brenda can fill her pool in 30 hours using the normal intake pipe. She can instead fill the pool in 45 hours using the garden hose. How long will it take to fill the empty pool if both the pipe and garden hose are running? |
We have a super fun division math game for kids! Long division is actually fun when kids understand the concepts and can get some low stress practice. Making a game out of long division is the perfect solution and this is an easy homemade math game perfect for your third or fourth graders learning long division.
## Long Division Math Game for Kids
I have really cherished memories of my dad helping me with math skills when I was a kid. He made learning into a game. It can be done, I promise!
Related: Math worksheets for kids
The mere mention of long division may make your kid cringe. But it’s an important fourth grade math skill and your kid *can* do it. Here’s a creative way to help them out with division, or make them faster at it.
The key to conquering long division is lots of repetition and practice, but that doesn’t have to mean all worksheets and scratch paper. Make this hands-on game, and explore the fun side of division.
## How to Play the Division Game
Have your child grab the pencil and a sheet of lined paper and review his multiplication facts by completing the following problems:
• 3 x 2=____
• 4 x 5 = ____
• 6 x 7= ____
• 8 x 9 = ____
• 9 x 9 = ____
• 7 x 4=____
• 8 x 3 = ____
• 5 x 8 = ____
• 9 x 5 = ____
• 10 x 10 = ____
When he’s done, check his answers. If he’s a little shaky on certain multiplication facts, it doesn’t hurt to try a few more problems.
## Time to Make the Math Game
### Step 1
When he’s ready, create the number tiles by writing the following numbers and symbols on the index cards.
• 2 sets of numbers 0-9
• 1 division sign
• 1 equal sign
• 1 set of numbers 10-100
• 1 decimal point
### Step 2
Cut out the tiles and attach the magnetic roll to the back of them.
### Step 3
Use the baking sheet and magnetic numbers to complete the following division problems:
• 250 / 2
• 1075 / 50
• 6728 / 46
• 9258 / 71
• 3478 / 62
• 8120 / 89
• 9671 / 34
• 754 / 12
Remind him to use the decimal point when needed.
When he’s done, help him check his answers.
As he gets more confident, encourage him to try problems with larger numbers.
No problem: just make plus sign, minus sign, and multiplication signs, too. And don’t miss even more math games for kids.
## Free Math Printables
These spring math printables encourage your children to practice addition and subtraction while getting them excited for the upcoming spring season! Does your child love dinosaurs? Then they’ll have a blast with this printable dinosaur themed math puzzle hunt game. Or for something more sweet, print out these Valentine’s day math worksheets!
Related: More math fun with place value games & math games
## Want More Educational Games? Check Out These Posts:
Welcome to Kids Activities!
My name is Holly Homer & I am the Dallas mom of three boys…
## 1 Comment
1. Raymond says:
I have really cherished memories of my dad helping me with math skills when I was a kid. He made learning into a game. It can be done, I promise! |
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# Why do psychiatrists ask you to count backwards?
## Why do psychiatrists ask you to count backwards?
An abnormal attention span can indicate attention deficit disorder (ADD), as well as a wide range of other difficulties. Your examiner may ask you to count backward from a certain number or spell a short word both forward and backward. You may also be asked to follow spoken instructions.
## How do you skip a count by 1?
To skip count you add the same number over and over. You can start at any number. When you count normally (like 1,2,3,4,5,6) you add 1 to get the next number. To count by 2s, you add 2 to get the next number.
## What is the purpose of skip counting?
Skip counting is also essential as it lays a mathematical foundation for developing a students ability on other mathematical skills. “Counting forwards and backwards in ones, or even in twos, fives and other multiples, are strategies that may be used to solve addition, subtraction, multiplication and division problems.
## How do you explain skip counting by 5?
The concept on skip counting by 5’s or fives is an essential skill to learn when making the jump from counting to basic addition. The sequence chart will help us to write the number to complete the series which involves skip counting by fives up to 20 times.
## How do you count by fives?
You can count by fives by adding five to the previous count. If you start with a zero then each number will end in either 5 or 0. The numbers that you would count if you started with 0 and counted by fives would be: 0, 5, 10, 15, 20, 25, 30, 35, 40, 45, 50, 55, 60, 65, 70, 75, 80, 85, 90, 95, 100.
## How is counting by 5s like counting by 10s?
When we count by 5 we say some of the same numbers as when we count by 10 (like 10, 20, 30, etc). They are both faster ways of counting than counting by 1. Counting by 10 is faster than counting by 5.
## Why is counting by 10 faster than counting by ones?
Counting in 10s is faster than counting in ones because counting in 10s gets you to larger numbers faster because you are going by larger numbers.
## How do you explain counting by 10s?
When you count by tens the numbers create a pattern. All the numbers end with a zero. The first digits are just like the numbers when you count (1, 2, 3, 4, 5, etc.). This pattern gives the numbers 10, 20, 30, 40, 50, etc.
Category: Uncategorized
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# Science:Infinite Series Module/Units/Unit 1/1.1 Infinite Sequences/1.1.11 Example rn
## Example
Determine the values of r so that the sequence
${\displaystyle \{r^{n}\}_{n=1}^{\infty }}$
is convergent.
## Complete Solution
We can solve this problem by considering cases for the value of r.
### Case 1
If r > 1, then rn tends to infinity as n tends to infinity. The sequence is divergent in this case.
### Case 2
If r= 1, then
${\displaystyle \lim _{n\rightarrow a}r^{n}=1}$
so the sequence is convergent for this case.
### Case 3
If ${\displaystyle -1, then
${\displaystyle \lim _{n\rightarrow \infty }r^{n}=0,}$
so the sequence is convergent for this case.
### Case 4
If ${\displaystyle r=-1}$, then
${\displaystyle \lim _{n\rightarrow \infty }(-1)^{n}}$
does not exist, so the sequence is divergent for this case.
### Case 5
If ${\displaystyle r<-1}$, then ${\displaystyle r^{n}}$ tends to negative infinity as ${\displaystyle n}$ does not tend to a single finite number. The sequence is divergent in this case.
### Summary
Therefore, the sequence ${\displaystyle \{r^{n}\}_{n=1}^{\infty }}$ is convergent when ${\displaystyle -1.
## Explanation of Each Step
### Case 1
Consider the case when ${\displaystyle r=2}$. Then our sequence becomes
${\displaystyle \{2,4,8,16,32,64,\ldots \}}$
which tends to infinity.
### Case 2
Here we are using a fundamental property of limits, that the limit of a constant equals that constant:
${\displaystyle \lim _{x\rightarrow \infty }c=c}$
for any constant ${\displaystyle c}$ and ${\displaystyle x\in \mathbb {R} }$.
### Case 3
Consider the case when ${\displaystyle r=1/2}$. Then our sequence becomes
${\displaystyle \{1/2,1/4,1/8,1/16,1/32,\ldots \}}$
which tends to zero.
Similarly, if ${\displaystyle r=-1/2}$. Then our sequence becomes
${\displaystyle \{-1/2,+1/4,-1/8,+1/16,-1/32,\ldots \}}$
which also tends to zero.
### Case 4
In this case, we have the sequence
${\displaystyle -1,+1,-1,+1,\ldots }$
As ${\displaystyle n}$ approaches infinity the sequence does not approach a unique value, so the limit does not exist.
### Case 5
This case is similar to Case 1. Consider the case when ${\displaystyle r=-2}$. Then our sequence becomes
${\displaystyle \{-2,+4,-8,+16,-32,+64,\ldots \}}$
The terms alternate between positive and negative numbers, and do not tend to a single finite number.
## Possible Challenge Areas
### Connecting Results to Definition of Convergence
In each of the cases, we used a limit to determine whether the sequence is convergent. According to our definition of convergence of a sequence, as long as our respective limits exist, then the sequence converges. |
## If x + y + z > 0, is z > 1 ?
##### This topic has expert replies
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### If x + y + z > 0, is z > 1 ?
by BTGModeratorVI » Thu Mar 19, 2020 5:27 am
If x + y + z > 0, is z > 1 ?
(1) z > x + y + 1
(2) x + y + 1 < 0
Source: Official Guide
### GMAT/MBA Expert
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### Re: If x + y + z > 0, is z > 1 ?
by [email protected] » Sat Mar 21, 2020 5:59 am
BTGModeratorVI wrote:
Thu Mar 19, 2020 5:27 am
If x + y + z > 0, is z > 1 ?
(1) z > x + y + 1
(2) x + y + 1 < 0
Source: Official Guide
Target question: Is z > 1
Given: x + y + z > 0
Statement 1: z > x + y +1
Let's create a similar inequality to x + y + z > 0
Take z > x + y +1 and subtract x and y from both sides to get: z - x - y > 1
We now have two inequalities with the inequality signs facing the same direction.
z - x - y > 1
x + y + z > 0
ADD them to get: 2z > 1
Divide both sides by 2 to get: z > 1/2
So, z COULD equal 2, in which case z > 1
Or z COULD equal 3/4, in which case z < 1
Since we cannot answer the target question with certainty, statement 1 is NOT SUFFICIENT
Statement 2: x + y + 1 < 0
Let's use the same strategy.
This time, let's multiply both sides by -1 to get: -x - y - 1 > 0
We now have two inequalities with the inequality signs facing the same direction.
-x - y - 1 > 0
x + y + z > 0
ADD them to get: z - 1 > 0
Add 1 to both sides to get z > 1
Perfect!!!
Since we can answer the target question with certainty, statement 2 is SUFFICIENT
Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
Newbie | Next Rank: 10 Posts
Posts: 9
Joined: 14 Jun 2019
### Re: If x + y + z > 0, is z > 1 ?
by gmatbyexample » Tue Aug 17, 2021 12:10 pm
Great solution above.
I had prepared a YouTube video for the same in case you are looking for a visual:
https://youtu.be/km49QKnuyoo
GMAT/MBA Coach (MBA, Columbia Business School)
https://GMATByExample.com |
# Here are the first five terms of a sequence. 30, 29, 27, 24. 20, ... Find, in terms of n, an expression for the nth term of the sequence.How to do this? Please explain clearly...! Thank you
hala718 | High School Teacher | (Level 1) Educator Emeritus
Posted on
30 , 29 , 27, 24, 20 , ...
Let us try and determine the difference between terms.\
a2-a1 = 20 - 30 = -1
a3- a2= 27- 29 = -2
a4- a3= 24- 27 = -3
a5- a4= 20 - 24 = -4
....
We notice that we do not have a constant difference between terms, However, we found that there is a sequence withthe difference between the terms.
Then,
a1= 30
a2= a1 - 1 = a1- 1(0)/2
a3= a2- 2 = (a1-2) -1 = a1- 3 = a1- 3(2)/2
a4= a3-3 = (a1-3) - 3= a1- 6 = a1- 4(3)/2
a5= a4-4 = (a1-6) - 4= a1- 10 = a1- 5(4)/2
a6= a5- 5 = (a1-10) - 5 = a1- 15 = a1- 6(5)/2
......
an = a1 - n(n-1)/2
neela | High School Teacher | (Level 3) Valedictorian
Posted on
The first 5 terms of the sequence are 30, 29,27,24,20.
So a1 = 30, a2 = 29, a3 = 27, a4 = 24, a5 = 20.
a2-a1 = -1, a3-a2 = -2, a4-a3 = -3, a5-a4 = -4...., an-an-1 = n-1.
Therefore an = a1 - 1 -2 -3 - ... -(n-1) = a1- {1+2+3+..(n-1)} = a1-n(n-1)/2.
an = 30-n(n-1)/2.
Therefore, an = 30-n(n-1)/2 is the expression for the nth term in terms of n.
Tally:
a1 = 30 -(1)(0)/2 = 30
a2 = 30 -2(2-1)/2 = 29
a3 = 30 - 3(3-1)/2 = 30 -3 = 27
a4 = 30 - 4(4-1)/2 = 30 - 6 = 24
a5 = 30 -5(5-1)/2 = 30 -10 = 20. |
# exponent properties
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Exponents is the power or degree to a given variable or number. The exponent can be any real number. There are many different properties of the exponents in algebra which help in solving many types of question having exponents. Mentioned below are some properties of exponents.
Multiplication rule: am * an = a(m+n) (Here the base is the same value a)
Division rule: am / an = a(m-n) (Here the base is the same value a)
Power of a power: (am)n = amn
Example 1: Find the value of x in the equation 3(x+2) = 27.
Solution: Here the given equation is 3(x+2) = 27.
We need to simplify the 27 further.
The number 27 can be written as 27 = 3* 3 * 3
So, 27 = 33
Now we get 3(x+2) = 33.
Since the base number is 3 we can equate the exponents.
X + 2 = 3 (subtracting 2 on both sides.)
X = 3 – 2.
Hence the value of x = 1.
Example 2: Find the x in the equation 102 = 1/100.
Solution: Here the given equation is 102 = 1/100.
The fraction, 1/100 = 100-1.
We need to simplify 100 here further.
The number 100 can be written as 100 = 10* 10
So, 100 = 102
Now we get 10(x) = (102)-1.
Using the power of power rule.
10(x) = (10-2)
Since the base number is 10 we can equate the exponents.
Hence the value of x = -2. |
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# Mathematics 334 Designed by Peter Nield Notes by Katherine Daignault
## Westwood High School Grade 9 Student
Basic Quadratic Function: y = x 2
parabola
vertex
Standard Form: y = a (x − h ) + k
2
“a”:
• If “a” is negative, function opens down.
• If “a” is positive, function opens up.
• If “a” is greater than 1, parabola gets skinnier.
• If “a” is less than 1, parabola gets wider.
“h”: Translates function horizontally.
“k”: Translates function vertically.
VERTEX = (h, k )
* “h” value always opposite sign!!
## Parabolas are symmetrical.
Axis of symmetry at x = h
EX:
y = 2( x + 3) − 1
2
x = -3
44
Mathematics 334 Designed by Peter Nield Notes by Katherine Daignault
Westwood High School Grade 9 Student
General Form: y = ax 2 + bx + c
“a”:
• If “a” is negative, function opens down.
• If “a” is positive, function opens up.
• If “a” is greater than 1, parabola gets skinnier.
• If “a” is less than 1, parabola gets wider.
“c”: the “y” intercept.
Converting Forms:
To convert general to standard form, you can complete the square.
EX:
y = 2 x 2 − 8 x + 12
(
y = 2 x 2 − 4x + 6 )
y = 2(x 2
− 4x + 4 − 4 + 6 )
[
y = 2 (x − 2) + 2
2
]
y = 2( x − 2) + 4
2
(
SHORTCUT: ax 2 + bx + c )
a=a
−b
h=
2a
4ac − b 2
k=
4a
EX:
y = 2 x 2 − 8 x + 12 a=2
b = −8
c = 12
y = a( x − h ) + k
2
a=a=2
−b 8
h= = =2
2a 4
4ac − b 2 4 × 2 × 12 − (− 8) 96 − 64
2
k= = = =4
4a 4× 2 8
y = 2( x − 2) + 4
2
45 |
Trigonometry – Tangent function (tan)
T-O-A
A tangent is like a special line that touches the edge of the circle at just one point and then continues away from the circle without curving. It doesn’t go into the circle or toward the center. Instead, it’s a straight line that just grazes the circle at one point.
The tangent function is one of the primary trigonometric functions and is used to relate the angles of a right triangle to the ratios of the lengths of its sides.
Here’s an explanation of the tangent function:
Tangent (tan θ):
• The tangent of an angle θ in a right triangle is defined as the ratio of the length of the side opposite to that angle to the length of the side adjacent to it.
• In simple terms, if you have a right triangle and you know one of the non-right angles (θ), you can find the tangent of that angle by taking the length of the side opposite θ and dividing it by the length of the side adjacent to θ.
The formula for the tangent function is:
In trigonometry, the development of trigonometric functions like the tangent function was a collaborative effort over many centuries. Early work on trigonometry can be traced back to Indian, Greek, and Islamic scholars. The tangent function, as we know it today, was refined and developed further in the European Renaissance, with contributions from mathematicians like Regiomontanus, Copernicus, and others.
Tangents and the tangent function are used in various real-life applications, particularly in fields that involve angles, slopes, and rates of change.
Engineers and architects use tangents when designing and constructing roads, bridges, and buildings. Tangents help in creating smooth transitions between curved and straight sections of roads or railroads. For example, when designing a highway, engineers use tangent lines to smoothly connect curved portions, ensuring that vehicles can transition from curves to straight sections without abrupt changes in direction. This not only improves safety but also provides a more comfortable driving experience.
Tangent lines are also essential in civil engineering for designing drainage systems, as they determine the slope at which water flows along roadways, helping to prevent flooding. |
# problems on hcf and lcm
6. The product of two numbers is 4107. If the H.C.F. of these numbers is 37, then the greater number is:
A. 101 B. 107 C. 111 D. 185
Explanation:
Let the numbers be 37a and 37b.
Then, 37a x 37b = 4107
ab = 3.
Now, co-primes with product 3 are (1, 3).
So, the required numbers are (37 x 1, 37 x 3) i.e., (37, 111).
Greater number = 111.
7. Three number are in the ratio of 3 : 4 : 5 and their L.C.M. is 2400. Their H.C.F. is:
A. 40 B. 80 C. 120 D. 200
Explanation:
Let the numbers be 3x, 4x and 5x.
Then, their L.C.M. = 60x.
So, 60x = 2400 or x = 40.
The numbers are (3 x 40), (4 x 40) and (5 x 40).
Hence, required H.C.F. = 40.
8. The G.C.D. of 1.08, 0.36 and 0.9 is:
A. 0.03 B. 0.9 C. 0.18 D. 0.108
Explanation:
Given numbers are 1.08, 0.36 and 0.90. H.C.F. of 108, 36 and 90 is 18,
H.C.F. of given numbers = 0.18.
9. The product of two numbers is 2028 and their H.C.F. is 13. The number of such pairs is:
A. 1 B. 2 C. 3 D. 4
Explanation:
Let the numbers 13a and 13b.
Then, 13a x 13b = 2028
ab = 12.
Now, the co-primes with product 12 are (1, 12) and (3, 4).
[Note: Two integers a and b are said to be coprime or relatively prime if they have no common positive factor other than 1 or, equivalently, if their greatest common divisor is 1 ]
So, the required numbers are (13 x 1, 13 x 12) and (13 x 3, 13 x 4).
Clearly, there are 2 such pairs.
10. The least multiple of 7, which leaves a remainder of 4, when divided by 6, 9, 15 and 18 is:
A. 74 B. 94 C. 184 D. 364 |
Properties of Algebra By: Will Bienkowski.
Presentation on theme: "Properties of Algebra By: Will Bienkowski."— Presentation transcript:
Properties of Algebra By: Will Bienkowski
Commutative Property Explanation Examples
The commutative property works for multiplication and addition. It states that switching the order of the numbers and variables will not change the sum or product. 6+3=3+6 is for addition. 8·6=6·8 is for multiplication
Associative Property Explanation Examples
The associative property works for multiplication and addition. It states that that if you switch the grouping of numbers or variables the sum or product will stay the same. 7+(6+2)=15 6+(7+2)=15 8(2·4)=64 2(8·4)=64
Identity Property Explanation Example
The identity property works for multiplication and addition. It states that anything added to zero or multiplied by one the result is that number. 9+0=9 6·1=6
Distributive Property
Explanation Examples The distributive property states that you can make hard and long problems easy and short. 8·104 8·100=800 8·4= =832
Equality Property Explanation Example
The equality property states that if you add or take away numbers from one side of the =,< or > you have to add or take the same number away from the other side. 3+b< b<6
Inverse Property Explanation Example
It states that addition and subtraction can undo each other and multiplication and division can undo each other. a-a=0 and a+(-a)=0 a\a=1 and a·(-a)=1 |
# Symmetric Matrix and Skew Symmetric Matrix: Properties of Symmetric Matrix (For CBSE, ICSE, IAS, NET, NRA 2022)
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To understand if a matrix is a symmetric matrix, it is very important to know about transpose of a matrix and how to find it. If we interchange rows and columns of an m × n matrix to get an n × m matrix, the new matrix is called the transpose of the given matrix. There are two possibilities for the number of rows (m) and columns (n) of a given matrix:
• If , the matrix is square
• If , the matrix is rectangular
For the second case, the transpose of a matrix can never be equal to it. This is because, for equality, the order of the matrices should be the same. Hence, the only case where the transpose of a matrix can be equal to it is when the matrix is square. But this is only the first condition. Even if the matrix is square, its transpose may or may not be equal to it. For example:
If then Here, we can see that
Let us take another example.
If we take the transpose of this matrix, we will get:
Whenever this happens for any matrix, that is whenever transpose of a matrix is equal to it, the matrix is known as a symmetric matrix. But how can we find whether a matrix is symmetric or not without finding its transpose? We know that:
If then (for all the values of i and j)
So, if for a matrix (for all the values of i and j) and then its transpose is equal to itself. A symmetric matrix will hence always be square. Some examples of symmetric matrices are:
## Properties of Symmetric Matrix
• Addition and difference of two symmetric matrices results in symmetric matrix.
• If A and B are two symmetric matrices and they follow the commutative property, i.e.. , then the product of A and B is symmetric.
• If matrix A is symmetric then and is also symmetric, where n is an integer.
• If A is a symmetric matrix then A-1 is also symmetric.
Skew Symmetric Matrix:
A matrix can be skew symmetric only if it is square. If the transpose of a matrix is equal to the negative of itself, the matrix is said to be skew symmetric. This means that for a matrix to be skew symmetric,
Also, for the matrix, (for all the values of i and j) . The diagonal elements of a skew symmetric matrix are equal to zero. This can be proved in following way:
The diagonal elements are characterized by the general formula,
, where
If then
If A is skew symmetric, then
So, , when (for all the values of i and j)
Some examples of skew symmetric matrices are:
## Properties of Skew Symmetric Matrix
• When we add two skew-symmetric matrices then the resultant matrix is also skew-symmetric.
• Scalar product of skew-symmetric matrix is also a skew-symmetric matrix.
• The diagonal of skew symmetric matrix consists of zero elements and therefore the sum of elements in the main diagonals is equal to zero.
• When identity matrix is added to skew symmetric matrix then the resultant matrix is invertible.
• The determinant of skew symmetric matrix is non-negative
Every square matrix can be expressed in the form of sum of a symmetric and a skew symmetric matrix
### How Do You Know if a Matrix is Symmetric?
To know if a matrix is symmetric, find the transpose of that matrix. If the transpose of that matrix is equal to itself, it is a symmetric matrix.
### Give an Example of a Matrix Which is Symmetric but Not Invertible
A zero (square) matrix is one such matrix which is clearly symmetric but not invertible.
### Is Symmetric Matrix Diagonalizable?
Yes, a symmetric matrix is always diagonalizable.
Example 1: Construct the matrix , where . State whether A is symmetric or skew – symmetric.
Solution:
From the given question, we come to know that we have to construct a matrix with 3 rows and 3 columns.
So, the matrix A with order is
Whether it is symmetric or skew symmetric matrix.
Hence, it is skew symmetric matrix.
Developed by: |
1. probability help..poisson process
During normal working hours, the help desk of a car repair company(also repairs small amount of motorbikes) receives telephone calls according to a Poisson process at the rate of 25 calls per hour.
it has been noted that 5% of calls relate to the company’s motorbike repairs.
Calculate the probability that exactly four calls are received between 10.20am and 10.30 am, exactly one of which relates to motorbike repairs.
Exactly eight calls are received between 10.20am and 10.30am one morning. Calculate the probability that exactly two of them relate to motorbike repairs
can any one show me how this is done please , thanks.
2. Originally Posted by frank567
During normal working hours, the help desk of a car repair company(also repairs small amount of motorbikes) receives telephone calls according to a Poisson process at the rate of 25 calls per hour.
it has been noted that 5% of calls relate to the company’s motorbike repairs.
Calculate the probability that exactly four calls are received between 10.20am and 10.30 am, exactly one of which relates to motorbike repairs.
Exactly eight calls are received between 10.20am and 10.30am one morning. Calculate the probability that exactly two of them relate to motorbike repairs
can any one show me how this is done please , thanks.
First to calculate $\displaystyle \lambda$ we need to find the average number of calls in a ten minute interval this gives
$\displaystyle \displaystyle \frac{25 \text{calls}}{1 \text{hr}}=\frac{25 \text{calls}}{1 \text{hr}}\frac{1 \text{ hr}}{60 \text{min}}=\frac{2.5 \text{ calls}}{10 \text{min}} \implies \lambda =2.5$
This gives the probability function
$\displaystyle f(k)=\frac{(2.5)^ke^{-2.5}}{k!}$
This gives $\displaystyle f(4) \approx .1336$
The 2nd part of this is a binomial distribution with
$\displaystyle g(k)=\binom{4}{k}(.05)^k(.95)^{4-k}$
This gives $\displaystyle g(1) \approx .1715$
So the probability of them both happening is
$\displaystyle f(4)\cdot g(1) \approx 0.0229$
Now you try the other one! |
# Relationship between lines and planes
### Line-Plane Intersecting
So both vectors are perpendicular to the given plane. calculate the cross product between the normal vector and the parallel vector from line. Lines and planes are perhaps the simplest of curves and surfaces in three dimensional space. They also will prove important as we seek to understand more. Shapes 5 Line & Plane Relationships Name_______________________ Worksheet A 1. Name all segments parallel to GE. 2. Name all segments parallel to.
Here, we describe that concept mathematically. Remember, the dot product of orthogonal vectors is zero.
This fact generates the vector equation of a plane: As described earlier in this section, any three points that do not all lie on the same line determine a plane. Given three such points, we can find an equation for the plane containing these points.
Solution To write an equation for a plane, we must find a normal vector for the plane. We start by identifying two vectors in the plane: We want to find the shortest distance from point P to the plane. Just as we find the two-dimensional distance between a point and a line by calculating the length of a line segment perpendicular to the line, we find the three-dimensional distance between a point and a plane by calculating the length of a line segment perpendicular to the plane.
When we describe the relationship between two planes in space, we have only two possibilities: When two planes are parallel, their normal vectors are parallel.
The intersection of two nonparallel planes is always a line. We can use the equations of the two planes to find parametric equations for the line of intersection. Solution Note that the two planes have nonparallel normals, so the planes intersect. Further, the origin satisfies each equation, so we know the line of intersection passes through the origin.
For example, builders constructing a house need to know the angle where different sections of the roof meet to know whether the roof will look good and drain properly. We can use normal vectors to calculate the angle between the two planes. We can do this because the angle between the normal vectors is the same as the angle between the planes.
The angle between two planes has the same measure as the angle between the normal vectors for the planes. Finding the Angle between Two Planes Determine whether each pair of planes is parallel, orthogonal, or neither.
If the planes are intersecting, but not orthogonal, find the measure of the angle between them. Give the answer in radians and round to two decimal places.
The normal vectors are parallel, so the planes are parallel. Hint Use the coefficients of the variables in each equation to find a normal vector for each plane. To find this distance, we simply select a point in one of the planes.
## Line–plane intersection
The distance from this point to the other plane is the distance between the planes. But, if the lines represent pipes in a chemical plant or tubes in an oil refinery or roads at an intersection of highways, confirming that the distance between them meets specifications can be both important and awkward to measure. One way is to model the two pipes as lines, using the techniques in this chapter, and then calculate the distance between them. The calculation involves forming vectors along the directions of the lines and using both the cross product and the dot product.
Industrial pipe installations often feature pipes running in different directions. How can we find the distance between two skew pipes?
Again, this can be done directly from the symmetric equations. Let's say I had a point, B, right over here. Well, notice the way I drew this, point A and B, they would define a line. For example, they would define this line right over here.
So they would define, they could define, this line right over here.
## Parallel and Perpendicular Lines and Planes
But both of these points and in fact, this entire line, exists on both of these planes that I just drew. And I could keep rotating these planes. I could have a plane that looks like this. I could have a plane that looks like this, that both of these points actually sit on. I'm essentially just rotating around this line that is defined by both of these points.
So two points does not seem to be sufficient. So there's no way that I could put-- Well, let's be careful here. So I could put a third point right over here, point C. And C sits on that line, and C sits on all of these planes. So it doesn't seem like just a random third point is sufficient to define, to pick out any one of these planes. But what if we make the constraint that the three points are not all on the same line.
Obviously, two points will always define a line. But what if the three points are not collinear.
Geometry Help from francinebavay.info - Point Line Plane
So instead of picking C as a point, what if we pick-- Is there any way to pick a point, D, that is not on this line, that is on more than one of these planes?
If I say, well, let's see, the point D-- Let's say point D is right over here. So it sits on this plane right over here, one of the first ones that I drew. So point D sits on that plane. Between point D, A, and B, there's only one plane that all three of those points sit on.
So a plane is defined by three non-colinear points.
So D, A, and B, you see, do not sit on the same line. A and B can sit on the same line.
### An introduction to geometry (Geometry, Points, Lines, Planes and Angles) – Mathplanet
D and A can sit on the same line. D and B can sit on the same line.
But A, B, and D does not sit on-- They are non-colinear. So for example, right over here in this diagram, we have a plane. |
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# Most important 5 Ratio questions sbi po
Government Jobs News
1. 4 years ago Rahul’s age was 3/4 times that of Ravi .Four years hence Rahul’s age will be 5/6 times
that of Ravi. What is the present age of Rahul?
1) 20 years 2) 15 years 3) 24 years 4) 16 years 5) None of these
Solution: (4). Let us consider present year as 2014 so 4 years ago is 2010 and 4 years hence from 2014
is 2018. So the gap between 2010 and 2018 is 8 years. So the equation becomes
(3x+8) / (4x+8) = 5/6 then x= 4
Present age of Rahul is 3x+4 = 3(4) + 4 = 16 years
2. Mr.Sundaram owned 950 gold coins all of which he distributed amongst his three sons Lakshman,
Arun and Nagaraj . Lakshman gave 25 gold coins to his wife, Arun donated 15 gold coins and Nagaraj made jewellery out of 30 gold coins. The new respective ratio of the coins left with them was 20:73:83. How many gold coins did Arun receive from Mr.Sundaram?
1) 350 2) 400 3) 380 4) 415 5)None of these
Solution: (3). Total coins spent by the sons is 70 (25+15+30).
So remaining coins is 880 (950 - 70) .
Given ratio is corresponding to remaining coins.
So 20x + 73x + 83x = 880. Then x = 5.
So Arun received 73x+15 = 73(5) + 15 = 380. So he received 380 gold coins.
3. Abi invested in three schemes A,B and C . The amounts in the Ratio of 2:3:4 respectively. If the
schemes offered interest at 20 p.c.p.a. , 16 p.c.p.a and 15 p.c.p.a. respectively .What will be respective ratio of the amounts after 1 year?
1) Cannot be determined 2) 10:8:5 3) 8:10:5 4) 15:14:12 5) None of these
Solution: (5). Let us consider the amount invested in three schemes A,B and C is 2x, 3x, 4x and assume
x = 100. Then they have 200,300,400.
As per interest rates Scheme A gets 20% which is 40 ( 200 x 20/100)
Similarly B gets 16% which is 48 ( 300 x 16/100)
Similarly C gets 15% which is 60 ( 400 x 15/100
So amount after 1 year is 240,348,460
then the ratio becomes 240:348:460 which is 60:87:115.
So answer is none of these
4. A sum of money divided among A, B, C and D in the ratio of 4:5:7:11 respectively .If the share of C is Rs.1351 then what is the total amount of money A and D together?
1) Rs. 2895 2) Rs.2316 3) Rs.2565 4) Rs.2123 5) None of these
Solution: (1). The ratio of A,B,C,D is 4x, 5x, 7x, 11x
It is given that share of C is 1351 then C = 7x = 1351, so solving x = 193.
Now question is A+D
Then A+D = 4x+11x = 15x = 15 x 193 = 2895. This is the share of A+D.
5. 38% of first number is 52% of second number .What is respective ratio of the first number to the
second number.
1) Cannot be determined 2) 16:9 3) 5:4 4) 26:19 5) None of these
Solution: (4) .let us consider first number as N1 and second number as N2
It is given 38% of N1 = 52% of N2
38% x N1 = 52% x N2
N1 / N2 = 52% / 38%
N1 / N2 = 52/38 further solving 26/19.
So N1:N2 = 26:19 |
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